Check if two nodes are in same subtree of the root node

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Given a Binary Tree with distinct nodes. Given two nodes node1 and node2, check if the two nodes lies in the same subtree of the root node. That is, either of the left and right subtrees of the root node.

For Example: In the above binary tree, node 3 and 8 are in the same subtree but 4 and 5 are in different subtree.

Prerequisite: Check if a node exist in Binary Tree.

The idea is similar of searching a node in Binary Tree. There are four different cases:

If both node1 and node2 are in left subtree of root node.

If both node1 and node2 are in right subtree of the root node.

If node1 is in the left subtree of the root node and node2 is in the right subtree of root node.

If node1 is in the right subtree of the root node and node2 is in the left subtree of root node.

Use the approach of searching a node in Binary Tree and check if any of the first two cases listed above is True. If any of the first two cases listed above is found True then print YES otherwise print NO.

Below is the implementation of the above approach:

C++

// C++ program to check if two nodes
// are in same subtrees of the root node
 
#include <iostream>
using namespace std;
 
// Binary tree node
struct Node {
    int data;
    struct Node *left, *right;
    Node(int data)
    {
        this->data = data;
        left = right = NULL;
    }
};
 
// Function to traverse the tree in preorder
// and check if the given node exists in
// a binary tree
bool ifNodeExists(struct Node* node, int key)
{
    if (node == NULL)
        return false;
 
    if (node->data == key)
        return true;
 
    /* then recur on left subtree */
    bool res1 = ifNodeExists(node->left, key);
 
    /* now recur on right subtree */
    bool res2 = ifNodeExists(node->right, key);
 
    return res1 || res2;
}
 
// Function to check if the two given nodes
// are in same subtrees of the root node
bool ifSameSubTree(Node* root, int node1, int node2)
{
    if (root == NULL)
       return false;
 
    // CASE 1: If both nodes are in left subtree
    if (ifNodeExists(root->left, node1)
        && ifNodeExists(root->left, node2)) {
        return true;
    }
 
    // CASE 2: If both nodes are in right subtree
    else if (ifNodeExists(root->right, node1)
             && ifNodeExists(root->right, node2)) {
        return true;
    }
 
    // CASE 3 and 4: Nodes are in different subtrees
    else
        return false;
}
 
// Driver Code
int main()
{
    struct Node* root = new Node(0);
    root->left = new Node(1);
    root->left->left = new Node(3);
    root->left->left->left = new Node(7);
    root->left->right = new Node(4);
    root->left->right->left = new Node(8);
    root->left->right->right = new Node(9);
    root->right = new Node(2);
    root->right->left = new Node(5);
    root->right->right = new Node(6);
 
    int node1 = 3, node2 = 8;
 
    if (ifSameSubTree(root, node1, node2))
        cout << "YES";
    else
        cout << "NO";
 
    return 0;
}

Java

// Java program to check if two nodes 
// are in same subtrees of the root node 
import java.util.*;
class solution
{
 
// Binary tree node 
 static class Node { 
    int data; 
     Node left, right; 
    Node(int data) 
    { 
        this.data = data; 
        left = right = null; 
    } 
} 
 
// Function to traverse the tree in preorder 
// and check if the given node exists in 
// a binary tree 
static boolean ifNodeExists( Node node, int key) 
{ 
    if (node == null) 
        return false; 
 
    if (node.data == key) 
        return true; 
 
    /* then recur on left subtree */
    boolean res1 = ifNodeExists(node.left, key); 
 
    /* now recur on right subtree */
    boolean res2 = ifNodeExists(node.right, key); 
 
    return res1 || res2; 
} 
 
// Function to check if the two given nodes 
// are in same subtrees of the root node 
static boolean ifSameSubTree(Node root, int node1, int node2) 
{ 
    if (root == null) 
    return false; 
 
    // CASE 1: If both nodes are in left subtree 
    if (ifNodeExists(root.left, node1) 
        && ifNodeExists(root.left, node2)) { 
        return true; 
    } 
 
    // CASE 2: If both nodes are in right subtree 
    else if (ifNodeExists(root.right, node1) 
            && ifNodeExists(root.right, node2)) { 
        return true; 
    } 
 
    // CASE 3 and 4: Nodes are in different subtrees 
    else
        return false; 
} 
 
// Driver Code 
public static void main(String args[])
{ 
     Node root = new Node(0); 
    root.left = new Node(1); 
    root.left.left = new Node(3); 
    root.left.left.left = new Node(7); 
    root.left.right = new Node(4); 
    root.left.right.left = new Node(8); 
    root.left.right.right = new Node(9); 
    root.right = new Node(2); 
    root.right.left = new Node(5); 
    root.right.right = new Node(6); 
 
    int node1 = 3, node2 = 8; 
 
    if (ifSameSubTree(root, node1, node2)) 
        System.out.println("YES"); 
    else
        System.out.println( "NO"); 
 
} 
}
//contributed by Arnab Kundu

Python3

"""Python3 program to check if two nodes 
are in same subtrees of the root node """
 
# A Binary Tree Node 
# Utility function to create a 
# new tree node 
class newNode: 
 
    # Constructor to create a newNode 
    def __init__(self, data): 
        self.data= data 
        self.left = None
        self.right = None
        self.visited = False
 
# Function to traverse the tree in 
# preorder and check if the given 
# node exists in a binary tree 
def ifNodeExists(node, key) :
    if (node == None):
        return False
 
    if (node.data == key):
        return True
 
    """ then recur on left subtree """
    res1 = ifNodeExists(node.left, key) 
 
    """ now recur on right subtree """
    res2 = ifNodeExists(node.right, key) 
 
    return res1 or res2 
 
# Function to check if the two given nodes 
# are in same subtrees of the root node 
def ifSameSubTree(root, node1, node2):
 
    if (root == None) :
        return False
 
    # CASE 1: If both nodes are in
    # left subtree 
    if (ifNodeExists(root.left, node1) 
        and ifNodeExists(root.left, node2)):
        return True
 
    # CASE 2: If both nodes are in 
    # right subtree 
    elif (ifNodeExists(root.right, node1) 
            and ifNodeExists(root.right, node2)):
        return True
     
    # CASE 3 and 4: Nodes are in 
    # different subtrees 
    else:
        return False
                         
# Driver Code
if __name__ == '__main__':
    root = newNode(0) 
    root.left = newNode(1) 
    root.left.left = newNode(3) 
    root.left.left.left = newNode(7) 
    root.left.right = newNode(4) 
    root.left.right.left = newNode(8) 
    root.left.right.right = newNode(9) 
    root.right = newNode(2) 
    root.right.left = newNode(5) 
    root.right.right = newNode(6) 
 
    node1 = 3
    node2 = 8
 
    if (ifSameSubTree(root, node1, node2)):
        print("YES") 
    else:
        print("NO")
 
# This code is contributed by 
# SHUBHAMSINGH10

C#

// C# program to check if two nodes 
// are in same subtrees of the root node 
using System;
 
class GFG
{ 
 
// Binary tree node 
class Node
{ 
    public int data; 
    public Node left, right; 
    public Node(int data) 
    { 
        this.data = data; 
        left = right = null; 
    } 
} 
 
// Function to traverse the tree in preorder 
// and check if the given node exists in 
// a binary tree 
static bool ifNodeExists( Node node, int key) 
{ 
    if (node == null) 
        return false; 
 
    if (node.data == key) 
        return true; 
 
    /* then recur on left subtree */
    bool res1 = ifNodeExists(node.left, key); 
 
    /* now recur on right subtree */
    bool res2 = ifNodeExists(node.right, key); 
 
    return res1 || res2; 
} 
 
// Function to check if the two given nodes 
// are in same subtrees of the root node 
static bool ifSameSubTree(Node root,
                int node1, int node2) 
{ 
    if (root == null) 
    return false; 
 
    // CASE 1: If both nodes are in left subtree 
    if (ifNodeExists(root.left, node1) 
        && ifNodeExists(root.left, node2)) 
    { 
        return true; 
    } 
 
    // CASE 2: If both nodes are in right subtree 
    else if (ifNodeExists(root.right, node1) 
            && ifNodeExists(root.right, node2)) 
    { 
        return true; 
    } 
 
    // CASE 3 and 4: Nodes are in different subtrees 
    else
        return false; 
} 
 
// Driver Code 
public static void Main() 
{ 
    Node root = new Node(0); 
    root.left = new Node(1); 
    root.left.left = new Node(3); 
    root.left.left.left = new Node(7); 
    root.left.right = new Node(4); 
    root.left.right.left = new Node(8); 
    root.left.right.right = new Node(9); 
    root.right = new Node(2); 
    root.right.left = new Node(5); 
    root.right.right = new Node(6); 
 
    int node1 = 3, node2 = 8; 
 
    if (ifSameSubTree(root, node1, node2)) 
        Console.WriteLine("YES"); 
    else
        Console.WriteLine( "NO"); 
 
} 
} 
 
/* This code is contributed by Rajput-Ji*/

Javascript

<script>
 
// JavaScript program to check if two nodes 
// are in same subtrees of the root node 
 
    // Binary tree node
class Node {
    constructor(val) {
        this.data = val;
        this.left = null;
        this.right = null;
    }
}
 
    // Function to traverse the tree in preorder
    // and check if the given node exists in
    // a binary tree
    function ifNodeExists(node , key) {
        if (node == null)
            return false;
 
        if (node.data == key)
            return true;
 
        /* then recur on left subtree */
        var res1 = ifNodeExists(node.left, key);
 
        /* now recur on right subtree */
        var res2 = ifNodeExists(node.right, key);
 
        return res1 || res2;
    }
 
    // Function to check if the two given nodes
    // are in same subtrees of the root node
    function ifSameSubTree(root , node1 , node2) {
        if (root == null)
            return false;
 
        // CASE 1: If both nodes are in left subtree
        if (ifNodeExists(root.left, node1) && 
        ifNodeExists(root.left, node2)) {
            return true;
        }
 
        // CASE 2: If both nodes are in right subtree
        else if (ifNodeExists(root.right, node1) && 
        ifNodeExists(root.right, node2)) {
            return true;
        }
 
        // CASE 3 and 4: Nodes are in different subtrees
        else
            return false;
    }
 
    // Driver Code
     
        var root = new Node(0);
        root.left = new Node(1);
        root.left.left = new Node(3);
        root.left.left.left = new Node(7);
        root.left.right = new Node(4);
        root.left.right.left = new Node(8);
        root.left.right.right = new Node(9);
        root.right = new Node(2);
        root.right.left = new Node(5);
        root.right.right = new Node(6);
 
        var node1 = 3, node2 = 8;
 
        if (ifSameSubTree(root, node1, node2))
            document.write("YES");
        else
            document.write("NO");
 
// This code is contributed by todaysgaurav
 
</script>

Output

YES

Complexity Analysis:

Time Complexity: O(N), as we are using recursion to traverse N nodes of the tree for checking if the nodes exists. Where N is the number of nodes in the tree.
Auxiliary Space: O(N), as though we are not using any explicit extra space but as we are using recursion a recursive call stack will be used which will cost O (N) space.

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