Print all pairs from two BSTs whose sum is greater than the given value

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Given two Binary Search Tree (BSTs) and a value X, the problem is to print all pairs from both the BSTs whose sum is greater than the given value X.\

Examples:

Input: 
BST 1:
                  5        
                /   \      
               3     7      
              / \   / \    
             2  4  6   8   
BST 2:
                 10        
                /   \      
               6     15      
              / \   /  \    
             3  8  11  18
X = 20
Output: The pairs are:
        (3, 18)
        (4, 18)
        (5, 18)
        (6, 18)
        (7, 18)
        (8, 18)
        (6, 15)
        (7, 15)
        (8, 15)

Naive Approach: For each node value A in BST 1, search the value in BST 2 which is greater than the (X – A). If the value is found then print the pair.

Below is the implementation of the approach:

C++

// C++ implementation to print pairs
// from two BSTs whose sum is greater
// the given value x
 
#include <bits/stdc++.h>
using namespace std;
 
// Structure of each node of BST
struct node {
    int key;
    struct node *left, *right;
};
 
// Function to create a new BST node
node* newNode(int item)
{
    node* temp = new node();
    temp->key = item;
    temp->left = temp->right = NULL;
    return temp;
}
 
// A utility function to insert a
// new node with given key in BST
struct node* insert(struct node* node, int key)
{
    // If the tree is empty, return a new node
    if (node == NULL)
        return newNode(key);
 
    // Otherwise, recur down the tree
    if (key < node->key)
        node->left = insert(node->left, key);
    else if (key > node->key)
        node->right = insert(node->right, key);
 
    // Return the (unchanged) node pointer
    return node;
}
 
// Function to return the size of
// the tree
int sizeOfTree(node* root)
{
    if (root == NULL) {
        return 0;
    }
 
    // Calculate left size recursively
    int left = sizeOfTree(root->left);
 
    // Calculate right size recursively
    int right = sizeOfTree(root->right);
 
    // Return total size recursively
    return (left + right + 1);
}
 
// Function to store inorder
// traversal of BST
void storeInorder(node* root, int inOrder[], int& index)
{
    // Base condition
    if (root == NULL) {
        return;
    }
 
    // Left recursive call
    storeInorder(root->left, inOrder, index);
 
    // Store elements in inorder array
    inOrder[index++] = root->key;
 
    // Right recursive call
    storeInorder(root->right, inOrder, index);
}
 
// Utility function to check the
// pair of BSTs whose sum is
// greater than given value x
void printPairUtil(int inOrder1[], int inOrder2[],
                   int index1, int index2, int k)
{
    // loop through every pairs formed from
    // inOrder1 and inOrder2 elements
    for (int i = 0; i < index1; i++) {
        for (int j = 0; j < index2; j++) {
            // store sum of the pairs
            int sum = inOrder1[i] + inOrder2[j];
 
            // if sum comes out to be greater than X
            // print the pair
            if (sum > k)
                cout << "(" << inOrder1[i] << ", "
                     << inOrder2[j] << ")" << endl;
        }
    }
}
 
// Function to check the
// pair of BSTs whose sum is
// greater than given value x
void printPairs(node* root1, node* root2, int k)
{
    // Store the size of BST1
    int numNode = sizeOfTree(root1);
 
    // Take auxiliary array for storing
    // The inorder traversal of BST1
    int inOrder1[numNode + 1];
    int index1 = 0;
 
    // Store the size of BST2
    numNode = sizeOfTree(root2);
 
    // Take auxiliary array for storing
    // The inorder traversal of BST2
    int inOrder2[numNode + 1];
    int index2 = 0;
 
    // Function call for storing
    // inorder traversal of BST1
    storeInorder(root1, inOrder1, index1);
 
    // Function call for storing
    // inorder traversal of BST1
    storeInorder(root2, inOrder2, index2);
 
    // Utility function call to count
    // the pair
    printPairUtil(inOrder1, inOrder2, index1, index2, k);
}
 
// Driver code
int main()
{
 
    /* Formation of BST 1
                    5
            / \
            3     7
            / \ / \
            2 4 6 8
    */
 
    struct node* root1 = NULL;
    root1 = insert(root1, 5);
    insert(root1, 3);
    insert(root1, 2);
    insert(root1, 4);
    insert(root1, 7);
    insert(root1, 6);
    insert(root1, 8);
 
    /* Formation of BST 2
                    10
            / \
            6     15
            / \ / \
            3 8 11 18
    */
 
    struct node* root2 = NULL;
    root2 = insert(root2, 10);
    insert(root2, 6);
    insert(root2, 15);
    insert(root2, 3);
    insert(root2, 8);
    insert(root2, 11);
    insert(root2, 18);
 
    int x = 20;
 
    // Print pairs
    printPairs(root1, root2, x);
 
    return 0;
}

Java

// Java implementation to print pairs
// from two BSTs whose sum is greater
// the given value x
 
// Structure of each Node of BST
class Node {
    int key;
    Node left, right;
}
 
class RefInteger {
    Integer value;
 
    public RefInteger(Integer value) { this.value = value; }
}
 
class GFG {
    // Function to create a new BST Node
    static Node newNode(int item)
    {
        Node temp = new Node();
        temp.key = item;
        temp.left = temp.right = null;
        return temp;
    }
 
    // A utility function to insert a
    // new Node with given key in BST
    static Node insert(Node node, int key)
    {
        // If the tree is empty,
        // return a new Node
        if (node == null)
            return newNode(key);
 
        // Otherwise, recur down the tree
        if (key < node.key)
            node.left = insert(node.left, key);
        else if (key > node.key)
            node.right = insert(node.right, key);
 
        // Return the (unchanged) Node pointer
        return node;
    }
 
    // Function to return the size of
    // the tree
    static int sizeOfTree(Node root)
    {
        if (root == null) {
            return 0;
        }
 
        // Calculate left size recursively
        int left = sizeOfTree(root.left);
 
        // Calculate right size recursively
        int right = sizeOfTree(root.right);
 
        // Return total size recursively
        return (left + right + 1);
    }
 
    // Function to store inorder
    // traversal of BST
    static void storeInorder(Node root, int inOrder[],
                             RefInteger index)
    {
        // Base condition
        if (root == null) {
            return;
        }
 
        // Left recursive call
        storeInorder(root.left, inOrder, index);
 
        // Store elements in inorder array
        inOrder[index.value++] = root.key;
 
        // Right recursive call
        storeInorder(root.right, inOrder, index);
    }
 
    // Utility function to check the
    // pair of BSTs whose sum is
    // greater than given value x
    static void printPairUtil(int inOrder1[],
                              int inOrder2[], int index1,
                              int index2, int k)
    {
        // loop through every pairs formed from inOrder1 and
        // inOrder2 elements
        for (int i = 0; i < index1; i++) {
            for (int j = 0; j < index2; j++) {
                // store sum of the pairs
                int sum = inOrder1[i] + inOrder2[j];
 
                // if sum comes out to be greater than k,
                // print the pair
                if (sum > k)
                    System.out.println("(" + inOrder1[i]
                                       + ", " + inOrder2[j]
                                       + ")");
            }
        }
    }
 
    // Function to check the pair of
    // BSTs whose sum is greater than
    // given value x
    static void printPairs(Node root1, Node root2, int k)
    {
        // Store the size of BST1
        int numNode = sizeOfTree(root1);
 
        // Take auxiliary array for storing
        // The inorder traversal of BST1
        int[] inOrder1 = new int[numNode + 1];
        RefInteger index1 = new RefInteger(0);
 
        // Store the size of BST2
        numNode = sizeOfTree(root2);
 
        // Take auxiliary array for storing
        // The inorder traversal of BST2
        int[] inOrder2 = new int[numNode + 1];
        RefInteger index2 = new RefInteger(0);
 
        // Function call for storing
        // inorder traversal of BST1
        storeInorder(root1, inOrder1, index1);
 
        // Function call for storing
        // inorder traversal of BST1
        storeInorder(root2, inOrder2, index2);
 
        // Utility function call to count
        // the pair
        printPairUtil(inOrder1, inOrder2, index1.value,
                      index2.value, k);
    }
 
    // Driver code
    public static void main(String[] args)
    {
 
        /* Formation of BST 1
                5
        / \
        3     7
        / \ / \
        2 4 6 8
        */
 
        Node root1 = null;
        root1 = insert(root1, 5);
        insert(root1, 3);
        insert(root1, 2);
        insert(root1, 4);
        insert(root1, 7);
        insert(root1, 6);
        insert(root1, 8);
 
        /* Formation of BST 2
                10
        / \
        6     15
        / \ / \
        3 8 11 18
        */
 
        Node root2 = null;
        root2 = insert(root2, 10);
        insert(root2, 6);
        insert(root2, 15);
        insert(root2, 3);
        insert(root2, 8);
        insert(root2, 11);
        insert(root2, 18);
 
        int x = 20;
 
        // Print pairs
        printPairs(root1, root2, x);
    }
}

Python3

# Python3 implementation to print pairs
# from two BSTs whose sum is greater
# the given value x
 
index = 0
 
# Structure of each node of BST
class newNode:
 
    def __init__(self, item):
 
        self.key = item
        self.left = None
        self.right = None
 
# A utility function to insert a
# new node with given key in BST
def insert(node, key):
 
    # If the tree is empty,
    # return a new node
    if (node == None):
        return newNode(key)
 
    # Otherwise, recur down the tree
    if (key < node.key):
        node.left = insert(node.left, key)
    elif (key > node.key):
        node.right = insert(node.right, key)
 
    # Return the (unchanged) node pointer
    return node
 
# Function to return the size of
# the tree
def sizeOfTree(root):
 
    if (root == None):
        return 0
 
    # Calculate left size recursively
    left = sizeOfTree(root.left)
 
    # Calculate right size recursively
    right = sizeOfTree(root.right)
 
    # Return total size recursively
    return (left + right + 1)
 
# Function to store inorder
# traversal of BST
def storeInorder(root, inOrder):
 
    global index
 
    # Base condition
    if (root == None):
        return
 
    # Left recursive call
    storeInorder(root.left, inOrder)
 
    # Store elements in inorder array
    inOrder[index] = root.key
    index += 1
 
    # Right recursive call
    storeInorder(root.right, inOrder)
 
# Utility function to check the
# pair of BSTs whose sum is
# greater than given value x
def printPairUtil(inOrder1, inOrder2, index1, index2, k):
     
    # loop through every pairs formed from
    # inOrder1 and inOrder2 elements
 
    for i in range(index1):
        for j in range(index2):
            # store sum of the pairs
            _sum = inOrder1[i] + inOrder2[j]
             
            # if sum comes out to be greater than X
            # print the pair
            if _sum > k:
                print("(", inOrder1[i], ",", inOrder2[j], ")")
           
 
# Function to check the
# pair of BSTs whose sum is
# greater than given value x
def printPairs(root1, root2, k):
     
    global index
     
    # Store the size of BST1
    numNode = sizeOfTree(root1)
 
    # Take auxiliary array for storing
    # The inorder traversal of BST1
    inOrder1 = [0 for i in range(numNode + 1)]
    index1 = 0
 
    # Store the size of BST2
    numNode = sizeOfTree(root2)
 
    # Take auxiliary array for storing
    # The inorder traversal of BST2
    inOrder2 = [0 for i in range(numNode + 1)]
    index2 = 0
 
    # Function call for storing
    # inorder traversal of BST1
    index = 0
    storeInorder(root1, inOrder1)
    temp1 = index
 
    # Function call for storing
    # inorder traversal of BST1
    index = 0
    storeInorder(root2, inOrder2)
    temp2 = index
 
    # Utility function call to count
    # the pair
    printPairUtil(inOrder1, inOrder2,
                temp1, temp2 , k)
 
# Driver code
if __name__ == '__main__':
     
    ''' Formation of BST 1
            5
        / \    
        3     7    
        / \ / \    
        2 4 6 8
    '''
 
    root1 = None
    root1 = insert(root1, 5)
    insert(root1, 3)
    insert(root1, 2)
    insert(root1, 4)
    insert(root1, 7)
    insert(root1, 6)
    insert(root1, 8)
     
    '''Formation of BST 2
            10
        / \    
        6     15    
        / \ / \    
        3 8 11 18
    '''
    root2 = None
    root2 = insert(root2, 10)
    insert(root2, 6)
    insert(root2, 15)
    insert(root2, 3)
    insert(root2, 8)
    insert(root2, 11)
    insert(root2, 18)
 
    x = 20
 
    # Print pairs
    printPairs(root1, root2, x)
     
# This code is contributed by Chandramani

C#

// C# implementation to print pairs
// from two BSTs whose sum is greater
// the given value x
 
using System;
 
class GFG{
 
public class Refint
{
    public int value;
 
    public Refint(int value)
    {
        this.value = value;
    }
}
 
// Structure of each Node of BST
public class Node
{
    public int key;
    public Node left, right;
};
 
// Function to create a new BST Node
static Node newNode(int item)
{
    Node temp = new Node();
    temp.key = item;
    temp.left = temp.right = null;
    return temp;
}
 
// A utility function to insert a
// new Node with given key in BST
static Node insert(Node Node, int key)
{
     
    // If the tree is empty,
    // return a new Node
    if (Node == null)
        return newNode(key);
 
    // Otherwise, recur down the tree
    if (key < Node.key)
        Node.left = insert(Node.left, key);
    else if (key > Node.key)
        Node.right = insert(Node.right, key);
 
    // Return the (unchanged) Node pointer
    return Node;
}
 
// Function to return the size of
// the tree
static int sizeOfTree(Node root)
{
    if (root == null)
    {
        return 0;
    }
 
    // Calculate left size recursively
    int left = sizeOfTree(root.left);
 
    // Calculate right size recursively
    int right = sizeOfTree(root.right);
 
    // Return total size recursively
    return (left + right + 1);
}
 
// Function to store inorder
// traversal of BST
static void storeInorder(Node root, int []inOrder,
                        Refint index)
{
     
    // Base condition
    if (root == null)
    {
        return;
    }
 
    // Left recursive call
    storeInorder(root.left, inOrder, index);
 
    // Store elements in inorder array
    inOrder[index.value++] = root.key;
 
    // Right recursive call
    storeInorder(root.right, inOrder, index);
}
 
// Function to print the pairs
static void print(int []inOrder1, int i,
                int index1, int value)
{
    while (i < index1)
    {
 
        Console.WriteLine("(" + inOrder1[i] +
                        ", " + value + ")");
        i++;
    }
}
 
// Utility function to check the
// pair of BSTs whose sum is
// greater than given value x
static void printPairUtil(int []inOrder1,
                        int []inOrder2,
                        int index1, int index2,
                        int k)
{
     // loop through every pairs formed from
    // inOrder1 and inOrder2 elements
    for (int i = 0; i < index1; i++) {
        for (int j = 0; j < index2; j++) {
            // store sum of the pairs
            int sum = inOrder1[i] + inOrder2[j];
  
            // if sum comes out to be greater than X
            // print the pair
            if (sum > k)
                Console.WriteLine("(" + inOrder1[i] +
                           ", " + inOrder2[j] + ")" );
        }
    }
}
 
// Function to check the pair of
// BSTs whose sum is greater than
// given value x
static void printPairs(Node root1,
                    Node root2, int k)
{
     
    // Store the size of BST1
    int numNode = sizeOfTree(root1);
 
    // Take auxiliary array for storing
    // The inorder traversal of BST1
    int[] inOrder1 = new int[numNode + 1];
    Refint index1 = new Refint(0);
 
    // Store the size of BST2
    numNode = sizeOfTree(root2);
 
    // Take auxiliary array for storing
    // The inorder traversal of BST2
    int[] inOrder2 = new int[numNode + 1];
    Refint index2 = new Refint(0);
 
    // Function call for storing
    // inorder traversal of BST1
    storeInorder(root1, inOrder1, index1);
 
    // Function call for storing
    // inorder traversal of BST1
    storeInorder(root2, inOrder2, index2);
 
    // Utility function call to count
    // the pair
    printPairUtil(inOrder1, inOrder2,
                index1.value,
                index2.value , k);
}
 
// Driver code
public static void Main(string[] args)
{
     
    /* Formation of BST 1
        5
    / \    
    3     7    
    / \ / \
    2 4 6 8
    */
 
    Node root1 = null;
    root1 = insert(root1, 5);
    insert(root1, 3);
    insert(root1, 2);
    insert(root1, 4);
    insert(root1, 7);
    insert(root1, 6);
    insert(root1, 8);
 
    /* Formation of BST 2
        10
    / \    
    6     15    
    / \ / \
    3 8 11 18
    */
 
    Node root2 = null;
    root2 = insert(root2, 10);
    insert(root2, 6);
    insert(root2, 15);
    insert(root2, 3);
    insert(root2, 8);
    insert(root2, 11);
    insert(root2, 18);
 
    int x = 20;
 
    // Print pairs
    printPairs(root1, root2, x);
}
}

Javascript

class Node {
    constructor(key) {
        this.key = key;
        this.left = null;
        this.right = null;
    }
}
 
// Function to insert a new node in BST
function insert(root, key) {
    if (root === null) {
        return new Node(key);
    }
 
    if (key < root.key) {
        root.left = insert(root.left, key);
    } else if (key > root.key) {
        root.right = insert(root.right, key);
    }
 
    return root;
}
 
// Function to calculate the size of the tree
function sizeOfTree(root) {
    if (root === null) {
        return 0;
    }
 
    let left = sizeOfTree(root.left);
    let right = sizeOfTree(root.right);
 
    return left + right + 1;
}
 
// Function to store inorder traversal of BST
function storeInorder(root, inOrder, index) {
    if (root === null) {
        return index;
    }
 
    index = storeInorder(root.left, inOrder, index);
    inOrder[index++] = root.key;
    index = storeInorder(root.right, inOrder, index);
 
    return index;
}
 
// Utility function to check pairs of BSTs whose sum is greater than x
function printPairUtil(inOrder1, inOrder2, index1, index2, k) {
    let result = [];
    let i = 0;
    let j = index2 - 1;
 
    while (i < index1 && j >= 0) {
        let sum = inOrder1[i] + inOrder2[j];
 
        if (sum > k) {
            for (let a = i; a < index1; a++) {
                for (let b = j; b >= 0; b--) {
                    let currSum = inOrder1[a] + inOrder2[b];
                    if (currSum > k) {
                        result.push(`(${inOrder1[a]}, ${inOrder2[b]})`);
                    } else {
                        break;
                    }
                }
            }
            j--;
        } else {
            i++;
        }
    }
 
    return result;
}
 
// Function to check pairs of BSTs whose sum is greater than x
function printPairs(root1, root2, k) {
    let numNode1 = sizeOfTree(root1);
    let inOrder1 = new Array(numNode1);
    let index1 = 0;
 
    let numNode2 = sizeOfTree(root2);
    let inOrder2 = new Array(numNode2);
    let index2 = 0;
 
    index1 = storeInorder(root1, inOrder1, index1);
    index2 = storeInorder(root2, inOrder2, index2);
 
    let pairs = printPairUtil(inOrder1, inOrder2, index1, index2, k);
    pairs.forEach(pair => console.log(pair));
}
 
 
let root1 = null;
root1 = insert(root1, 5);
insert(root1, 3);
insert(root1, 2);
insert(root1, 4);
insert(root1, 7);
insert(root1, 6);
insert(root1, 8);
 
let root2 = null;
root2 = insert(root2, 10);
insert(root2, 6);
insert(root2, 15);
insert(root2, 3);
insert(root2, 8);
insert(root2, 11);
insert(root2, 18);
 
let x = 20;
 
printPairs(root1, root2, x);

Output

(3, 18)
(4, 18)
(5, 18)
(6, 15)
(6, 18)
(7, 15)
(7, 18)
(8, 15)
(8, 18)

Time Complexity: O(N1*N2) where N1 and N2 are size of inOrder1 and inOrder2 arrays respectively as two nested loops are executed in printPairUtil function.

Space Complexity: O(N1+N2) as two arrays inOrder1 and inOrder2 are created.

Efficient Approach:

Traverse BST 1 from smallest value to node to largest by taking index i. This can be achieved with the help of inorder traversal.
Traverse BST 2 from largest value node to smallest by taking index j. This can be achieved with the help of inorder traversal.
Perform these two traversals one by one and store into two array.
Sum up the corresponding node’s value from both the BSTs at a particular instance of traversals.
- If sum > x, then print pair and decrement j by 1.
- If x > sum, then increment i by 1.

Below is the implementation of the above approach:

C++

// C++ implementation to print pairs
// from two BSTs whose sum is greater
// the given value x
 
#include <bits/stdc++.h>
using namespace std;
 
// Structure of each node of BST
struct node {
    int key;
    struct node *left, *right;
};
 
// Function to create a new BST node
node* newNode(int item)
{
    node* temp = new node();
    temp->key = item;
    temp->left = temp->right = NULL;
    return temp;
}
 
// A utility function to insert a
// new node with given key in BST
struct node* insert(struct node* node,
                    int key)
{
    // If the tree is empty, return a new node
    if (node == NULL)
        return newNode(key);
 
    // Otherwise, recur down the tree
    if (key < node->key)
        node->left = insert(node->left,
                            key);
    else if (key > node->key)
        node->right = insert(node->right,
                             key);
 
    // Return the (unchanged) node pointer
    return node;
}
 
// Function to return the size of
// the tree
int sizeOfTree(node* root)
{
    if (root == NULL) {
        return 0;
    }
 
    // Calculate left size recursively
    int left = sizeOfTree(root->left);
 
    // Calculate right size recursively
    int right = sizeOfTree(root->right);
 
    // Return total size recursively
    return (left + right + 1);
}
 
// Function to store inorder
// traversal of BST
void storeInorder(node* root,
                  int inOrder[],
                  int& index)
{
    // Base condition
    if (root == NULL) {
        return;
    }
 
    // Left recursive call
    storeInorder(root->left,
                 inOrder,
                 index);
 
    // Store elements in inorder array
    inOrder[index++] = root->key;
 
    // Right recursive call
    storeInorder(root->right,
                 inOrder,
                 index);
}
 
// Function to print the pairs
void print(int inOrder1[], int i,
           int index1, int value)
{
    while (i < index1) {
        cout << "(" << inOrder1[i]
             << ", " << value
             << ")" << endl;
        i++;
    }
}
 
// Utility function to check the
// pair of BSTs whose sum is
// greater than given value x
void printPairUtil(int inOrder1[],
                   int inOrder2[],
                   int index1,
                   int j, int k)
{
    int i = 0;
 
    while (i < index1 && j >= 0) {
 
        if (inOrder1[i] + inOrder2[j] > k) {
            print(inOrder1, i,
                  index1, inOrder2[j]);
            j--;
        }
        else {
            i++;
        }
    }
}
 
// Function to check the
// pair of BSTs whose sum is
// greater than given value x
void printPairs(node* root1,
                node* root2, int k)
{
    // Store the size of BST1
    int numNode = sizeOfTree(root1);
 
    // Take auxiliary array for storing
    // The inorder traversal of BST1
    int inOrder1[numNode + 1];
    int index1 = 0;
 
    // Store the size of BST2
    numNode = sizeOfTree(root2);
 
    // Take auxiliary array for storing
    // The inorder traversal of BST2
    int inOrder2[numNode + 1];
    int index2 = 0;
 
    // Function call for storing
    // inorder traversal of BST1
    storeInorder(root1, inOrder1,
                 index1);
 
    // Function call for storing
    // inorder traversal of BST1
    storeInorder(root2, inOrder2,
                 index2);
 
    // Utility function call to count
    // the pair
    printPairUtil(inOrder1, inOrder2,
                  index1, index2 - 1, k);
}
 
// Driver code
int main()
{
 
    /* Formation of BST 1 
             5
           /   \      
          3     7     
         / \   / \    
         2  4  6  8  
    */
 
    struct node* root1 = NULL;
    root1 = insert(root1, 5);
    insert(root1, 3);
    insert(root1, 2);
    insert(root1, 4);
    insert(root1, 7);
    insert(root1, 6);
    insert(root1, 8);
 
    /* Formation of BST 2 
            10
           /   \      
          6     15     
         / \   / \    
        3   8 11  18  
    */
 
    struct node* root2 = NULL;
    root2 = insert(root2, 10);
    insert(root2, 6);
    insert(root2, 15);
    insert(root2, 3);
    insert(root2, 8);
    insert(root2, 11);
    insert(root2, 18);
 
    int x = 20;
 
    // Print pairs
    printPairs(root1, root2, x);
 
    return 0;
}

Java

// Java implementation to print pairs
// from two BSTs whose sum is greater
// the given value x
class GFG{
 
static class RefInteger
{
    Integer value;
 
    public RefInteger(Integer value) 
    {
        this.value = value;
    }
}
 
// Structure of each Node of BST
static class Node 
{
    int key;
    Node left, right;
};
 
// Function to create a new BST Node
static Node newNode(int item) 
{
    Node temp = new Node();
    temp.key = item;
    temp.left = temp.right = null;
    return temp;
}
 
// A utility function to insert a
// new Node with given key in BST
static Node insert(Node Node, int key)
{
     
    // If the tree is empty, 
    // return a new Node
    if (Node == null)
        return newNode(key);
 
    // Otherwise, recur down the tree
    if (key < Node.key)
        Node.left = insert(Node.left, key);
    else if (key > Node.key)
        Node.right = insert(Node.right, key);
 
    // Return the (unchanged) Node pointer
    return Node;
}
 
// Function to return the size of
// the tree
static int sizeOfTree(Node root) 
{
    if (root == null) 
    {
        return 0;
    }
 
    // Calculate left size recursively
    int left = sizeOfTree(root.left);
 
    // Calculate right size recursively
    int right = sizeOfTree(root.right);
 
    // Return total size recursively
    return (left + right + 1);
}
 
// Function to store inorder
// traversal of BST
static void storeInorder(Node root, int inOrder[],
                         RefInteger index) 
{
     
    // Base condition
    if (root == null) 
    {
        return;
    }
 
    // Left recursive call
    storeInorder(root.left, inOrder, index);
 
    // Store elements in inorder array
    inOrder[index.value++] = root.key;
 
    // Right recursive call
    storeInorder(root.right, inOrder, index);
}
 
// Function to print the pairs
static void print(int inOrder1[], int i, 
                  int index1, int value) 
{
    while (i < index1) 
    {
        System.out.println("(" + inOrder1[i] + 
                           ", " + value + ")");
        i++;
    }
}
 
// Utility function to check the
// pair of BSTs whose sum is
// greater than given value x
static void printPairUtil(int inOrder1[], 
                          int inOrder2[], 
                          int index1, int j,
                          int k)
{
    int i = 0;
 
    while (i < index1 && j >= 0) 
    {
        if (inOrder1[i] + inOrder2[j] > k)
        {
            print(inOrder1, i, index1, 
                  inOrder2[j]);
 
            j--;
        }
        else
        {
            i++;
        }
    }
}
 
// Function to check the pair of 
// BSTs whose sum is greater than 
// given value x
static void printPairs(Node root1,
                       Node root2, int k)
{
     
    // Store the size of BST1
    int numNode = sizeOfTree(root1);
 
    // Take auxiliary array for storing
    // The inorder traversal of BST1
    int[] inOrder1 = new int[numNode + 1];
    RefInteger index1 = new RefInteger(0);
 
    // Store the size of BST2
    numNode = sizeOfTree(root2);
 
    // Take auxiliary array for storing
    // The inorder traversal of BST2
    int[] inOrder2 = new int[numNode + 1];
    RefInteger index2 = new RefInteger(0);
 
    // Function call for storing
    // inorder traversal of BST1
    storeInorder(root1, inOrder1, index1);
 
    // Function call for storing
    // inorder traversal of BST1
    storeInorder(root2, inOrder2, index2);
 
    // Utility function call to count
    // the pair
    printPairUtil(inOrder1, inOrder2, 
                  index1.value, 
                  index2.value - 1, k);
}
 
// Driver code
public static void main(String[] args) 
{
     
    /* Formation of BST 1 
         5
       /   \      
      3     7     
     / \   / \    
    2  4  6   8  
    */
 
    Node root1 = null;
    root1 = insert(root1, 5);
    insert(root1, 3);
    insert(root1, 2);
    insert(root1, 4);
    insert(root1, 7);
    insert(root1, 6);
    insert(root1, 8);
 
    /* Formation of BST 2 
        10
       /   \      
      6     15     
     / \   / \    
    3   8 11  18  
    */
 
    Node root2 = null;
    root2 = insert(root2, 10);
    insert(root2, 6);
    insert(root2, 15);
    insert(root2, 3);
    insert(root2, 8);
    insert(root2, 11);
    insert(root2, 18);
 
    int x = 20;
 
    // Print pairs
    printPairs(root1, root2, x);
}
}
 
// This code is contributed by sanjeev2552

Python3

# Python3 implementation to print pairs
# from two BSTs whose sum is greater
# the given value x
index = 0
 
# Structure of each node of BST
class newNode:
     
    def __init__(self, item):
         
        self.key = item
        self.left = None
        self.right = None
 
# A utility function to insert a
# new node with given key in BST
def insert(node, key):
     
    # If the tree is empty, 
    # return a new node
    if (node == None):
        return newNode(key)
 
    # Otherwise, recur down the tree
    if (key < node.key):
        node.left = insert(node.left, key)
    elif (key > node.key):
        node.right = insert(node.right, key)
 
    # Return the (unchanged) node pointer
    return node
 
# Function to return the size of
# the tree
def sizeOfTree(root):
     
    if (root == None):
        return 0
         
    # Calculate left size recursively
    left = sizeOfTree(root.left)
 
    # Calculate right size recursively
    right = sizeOfTree(root.right)
 
    # Return total size recursively
    return (left + right + 1)
 
# Function to store inorder
# traversal of BST
def storeInorder(root, inOrder):
     
    global index
     
    # Base condition
    if (root == None):
        return
 
    # Left recursive call
    storeInorder(root.left, inOrder)
 
    # Store elements in inorder array
    inOrder[index] = root.key
    index += 1
 
    # Right recursive call
    storeInorder(root.right, inOrder)
 
# Function to print the pairs
def print1(inOrder1, i, index1, value):
     
    while (i < index1):
        print("(", inOrder1[i], ",", value, ")")
        i += 1
 
# Utility function to check the
# pair of BSTs whose sum is
# greater than given value x
def printPairUtil(inOrder1, inOrder2,
                  index1, j, k):
                       
    i = 0
 
    while (i < index1 and j >= 0):
        if (inOrder1[i] + inOrder2[j] > k):
            print1(inOrder1, i, index1, inOrder2[j])
            j -= 1
        else:
            i += 1
 
# Function to check the
# pair of BSTs whose sum is
# greater than given value x
def printPairs(root1, root2, k):
     
    global index
     
    # Store the size of BST1
    numNode = sizeOfTree(root1)
 
    # Take auxiliary array for storing
    # The inorder traversal of BST1
    inOrder1 = [0 for i in range(numNode + 1)]
    index1 = 0
 
    # Store the size of BST2
    numNode = sizeOfTree(root2)
 
    # Take auxiliary array for storing
    # The inorder traversal of BST2
    inOrder2 = [0 for i in range(numNode + 1)]
    index2 = 0
 
    # Function call for storing
    # inorder traversal of BST1
    index = 0
    storeInorder(root1, inOrder1)
    temp1 = index
 
    # Function call for storing
    # inorder traversal of BST1
    index = 0
    storeInorder(root2, inOrder2)
    temp2 = index
 
    # Utility function call to count
    # the pair
    printPairUtil(inOrder1, inOrder2, 
                  temp1, temp2 - 1, k)
 
# Driver code
if __name__ == '__main__':
     
    ''' Formation of BST 1 
             5 
           /   \       
          3     7      
         / \   / \     
         2  4  6  8   
    '''
 
    root1 = None
    root1 = insert(root1, 5)
    insert(root1, 3)
    insert(root1, 2)
    insert(root1, 4)
    insert(root1, 7)
    insert(root1, 6)
    insert(root1, 8)
     
    '''Formation of BST 2 
            10 
           /   \       
          6     15      
         / \   / \     
        3   8 11  18   
    '''
    root2 = None
    root2 = insert(root2, 10)
    insert(root2, 6)
    insert(root2, 15)
    insert(root2, 3)
    insert(root2, 8)
    insert(root2, 11)
    insert(root2, 18)
 
    x = 20
 
    # Print pairs
    printPairs(root1, root2, x)
     
# This code is contributed by ipg2016107

C#

// C# implementation to print pairs
// from two BSTs whose sum is greater
// the given value x
 
using System;
 
class GFG{
  
public class Refint
{
    public int value;
  
    public Refint(int value) 
    {
        this.value = value;
    }
}
  
// Structure of each Node of BST
public class Node 
{
    public int key;
    public Node left, right;
};
  
// Function to create a new BST Node
static Node newNode(int item) 
{
    Node temp = new Node();
    temp.key = item;
    temp.left = temp.right = null;
    return temp;
}
  
// A utility function to insert a
// new Node with given key in BST
static Node insert(Node Node, int key)
{
      
    // If the tree is empty, 
    // return a new Node
    if (Node == null)
        return newNode(key);
  
    // Otherwise, recur down the tree
    if (key < Node.key)
        Node.left = insert(Node.left, key);
    else if (key > Node.key)
        Node.right = insert(Node.right, key);
  
    // Return the (unchanged) Node pointer
    return Node;
}
  
// Function to return the size of
// the tree
static int sizeOfTree(Node root) 
{
    if (root == null) 
    {
        return 0;
    }
  
    // Calculate left size recursively
    int left = sizeOfTree(root.left);
  
    // Calculate right size recursively
    int right = sizeOfTree(root.right);
  
    // Return total size recursively
    return (left + right + 1);
}
  
// Function to store inorder
// traversal of BST
static void storeInorder(Node root, int []inOrder,
                         Refint index) 
{
      
    // Base condition
    if (root == null) 
    {
        return;
    }
  
    // Left recursive call
    storeInorder(root.left, inOrder, index);
  
    // Store elements in inorder array
    inOrder[index.value++] = root.key;
  
    // Right recursive call
    storeInorder(root.right, inOrder, index);
}
  
// Function to print the pairs
static void print(int []inOrder1, int i, 
                  int index1, int value) 
{
    while (i < index1) 
    {
 
        Console.WriteLine("(" + inOrder1[i] + 
                           ", " + value + ")");
        i++;
    }
}
  
// Utility function to check the
// pair of BSTs whose sum is
// greater than given value x
static void printPairUtil(int []inOrder1, 
                          int []inOrder2, 
                          int index1, int j,
                          int k)
{
    int i = 0;
  
    while (i < index1 && j >= 0) 
    {
        if (inOrder1[i] + inOrder2[j] > k)
        {
            print(inOrder1, i, index1, 
                  inOrder2[j]);
  
            j--;
        }
        else
        {
            i++;
        }
    }
}
  
// Function to check the pair of 
// BSTs whose sum is greater than 
// given value x
static void printPairs(Node root1,
                       Node root2, int k)
{
      
    // Store the size of BST1
    int numNode = sizeOfTree(root1);
  
    // Take auxiliary array for storing
    // The inorder traversal of BST1
    int[] inOrder1 = new int[numNode + 1];
    Refint index1 = new Refint(0);
  
    // Store the size of BST2
    numNode = sizeOfTree(root2);
  
    // Take auxiliary array for storing
    // The inorder traversal of BST2
    int[] inOrder2 = new int[numNode + 1];
    Refint index2 = new Refint(0);
  
    // Function call for storing
    // inorder traversal of BST1
    storeInorder(root1, inOrder1, index1);
  
    // Function call for storing
    // inorder traversal of BST1
    storeInorder(root2, inOrder2, index2);
  
    // Utility function call to count
    // the pair
    printPairUtil(inOrder1, inOrder2, 
                  index1.value, 
                  index2.value - 1, k);
}
  
// Driver code
public static void Main(string[] args) 
{
      
    /* Formation of BST 1 
         5
       /   \      
      3     7     
     / \   / \    
    2  4  6   8  
    */
  
    Node root1 = null;
    root1 = insert(root1, 5);
    insert(root1, 3);
    insert(root1, 2);
    insert(root1, 4);
    insert(root1, 7);
    insert(root1, 6);
    insert(root1, 8);
  
    /* Formation of BST 2 
        10
       /   \      
      6     15     
     / \   / \    
    3   8 11  18  
    */
  
    Node root2 = null;
    root2 = insert(root2, 10);
    insert(root2, 6);
    insert(root2, 15);
    insert(root2, 3);
    insert(root2, 8);
    insert(root2, 11);
    insert(root2, 18);
  
    int x = 20;
  
    // Print pairs
    printPairs(root1, root2, x);
}
}
 
// This code is contributed by rutvik_56

Javascript

<script>
 
    // JavaScript implementation to print pairs
    // from two BSTs whose sum is greater
    // the given value x
     
    class Refint
    {
        constructor(value)
        {
            this.value = value;
        }
    }
 
    // Structure of each Node of BST
    class Node
    {
        constructor(item) {
           this.left = null;
           this.right = null;
           this.key = item;
        }
    }
 
    // Function to create a new BST Node
    function newNode(item)
    {
        let temp = new Node(item);
        return temp;
    }
 
    // A utility function to insert a
    // new Node with given key in BST
    function insert(Node, key)
    {
 
        // If the tree is empty,
        // return a new Node
        if (Node == null)
            return newNode(key);
 
        // Otherwise, recur down the tree
        if (key < Node.key)
            Node.left = insert(Node.left, key);
        else if (key > Node.key)
            Node.right = insert(Node.right, key);
 
        // Return the (unchanged) Node pointer
        return Node;
    }
 
    // Function to return the size of
    // the tree
    function sizeOfTree(root)
    {
        if (root == null)
        {
            return 0;
        }
 
        // Calculate left size recursively
        let left = sizeOfTree(root.left);
 
        // Calculate right size recursively
        let right = sizeOfTree(root.right);
 
        // Return total size recursively
        return (left + right + 1);
    }
 
    // Function to store inorder
    // traversal of BST
    function storeInorder(root, inOrder, index)
    {
 
        // Base condition
        if (root == null)
        {
            return;
        }
 
        // Left recursive call
        storeInorder(root.left, inOrder, index);
 
        // Store elements in inorder array
        inOrder[index.value++] = root.key;
 
        // Right recursive call
        storeInorder(root.right, inOrder, index);
    }
 
    // Function to print the pairs
    function print(inOrder1, i, index1, value)
    {
        while (i < index1)
        {
 
            document.write("(" + inOrder1[i] +
                               ", " + value + ")" + "</br>");
            i++;
        }
    }
 
    // Utility function to check the
    // pair of BSTs whose sum is
    // greater than given value x
    function printPairUtil(inOrder1, inOrder2, index1, j, k)
    {
        let i = 0;
 
        while (i < index1 && j >= 0)
        {
            if (inOrder1[i] + inOrder2[j] > k)
            {
                print(inOrder1, i, index1,
                      inOrder2[j]);
 
                j--;
            }
            else
            {
                i++;
            }
        }
    }
 
    // Function to check the pair of
    // BSTs whose sum is greater than
    // given value x
    function printPairs(root1, root2, k)
    {
 
        // Store the size of BST1
        let numNode = sizeOfTree(root1);
 
        // Take auxiliary array for storing
        // The inorder traversal of BST1
        let inOrder1 = new Array(numNode + 1);
        let index1 = new Refint(0);
 
        // Store the size of BST2
        numNode = sizeOfTree(root2);
 
        // Take auxiliary array for storing
        // The inorder traversal of BST2
        let inOrder2 = new Array(numNode + 1);
        let index2 = new Refint(0);
 
        // Function call for storing
        // inorder traversal of BST1
        storeInorder(root1, inOrder1, index1);
 
        // Function call for storing
        // inorder traversal of BST1
        storeInorder(root2, inOrder2, index2);
 
        // Utility function call to count
        // the pair
        printPairUtil(inOrder1, inOrder2,
                      index1.value,
                      index2.value - 1, k);
    }
     
    /* Formation of BST 1
         5
       /   \     
      3     7    
     / \   / \   
    2  4  6   8 
    */
   
    let root1 = null;
    root1 = insert(root1, 5);
    insert(root1, 3);
    insert(root1, 2);
    insert(root1, 4);
    insert(root1, 7);
    insert(root1, 6);
    insert(root1, 8);
   
    /* Formation of BST 2
        10
       /   \     
      6     15    
     / \   / \   
    3   8 11  18 
    */
   
    let root2 = null;
    root2 = insert(root2, 10);
    insert(root2, 6);
    insert(root2, 15);
    insert(root2, 3);
    insert(root2, 8);
    insert(root2, 11);
    insert(root2, 18);
   
    let x = 20;
   
    // Print pairs
    printPairs(root1, root2, x);
 
</script>

Output

(3, 18)
(4, 18)
(5, 18)
(6, 18)
(7, 18)
(8, 18)
(6, 15)
(7, 15)
(8, 15)

Time complexity: O(n1 * h2), where n1 is the number of nodes in the first BST and h2 is the height of the second BST.
Auxiliary Space: O(n), where n is the total number of nodes in the two BSTs

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