Count of root to leaf paths in a Binary Tree that form an AP

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Given a Binary Tree, the task is to count all paths from root to leaf which forms an Arithmetic Progression.

Examples:

Input:

Output: 2
Explanation:
The paths that form an AP in the given tree from root to leaf are:

1->3->5 (A.P. with common difference 2)

1->6->11 (A.P. with common difference 5)

Input:

Output: 1
Explanation:
The path that form an AP in the given tree from root to leaf is 1->10->19 (A.P. with difference 9)

Approach: The problem can be solved using the Preorder Traversal. Follow the steps below to solve the problem:

Perform Preorder Traversal on the given binary tree.
Initialize an array arr[] to store the path.
Initialize count = 0, to store the count of paths which forms an A.P.
After reaching the leaf node, check if the current elements in the array(i.e. the node values from root to leaf path) forms an A.P..
- If so, increment the count
- After the complete traversal of the tree, print the count.

Below is the implementation of above approach:

C++

// C++ implementation to count 
// the path which forms an A.P. 
#include <bits/stdc++.h> 
using namespace std; 
 
int count = 0; 
 
// Node structure 
struct Node { 
    int val; 
    // left and right child of the node 
    Node *left, *right; 
    // initialization constructor 
    Node(int x) 
    { 
        val = x; 
        left = NULL; 
        right = NULL; 
    } 
}; 
 
// Function to check if path 
// format A.P. or not 
bool check(vector<int> arr) 
{ 
 
    if (arr.size() == 1) 
        return true; 
 
    // if size of arr is greater than 2 
    int d = arr[1] - arr[0]; 
 
    for (int i = 2; i < arr.size(); i++) { 
        if (arr[i] - arr[i - 1] != d) 
            return false; 
    } 
 
    return true; 
} 
 
// Function to find the maximum 
// setbits sum from root to leaf 
int countAP(Node* root, vector<int> arr) 
{ 
    if (!root) 
        return 0; 
 
    arr.push_back(root->val); 
 
    // If the node is a leaf node 
    if (root->left == NULL 
        && root->right == NULL) { 
        if (check(arr)) 
            return 1; 
        return 0; 
    } 
 
    // Traverse left subtree 
    int x = countAP(root->left, arr); 
 
    // Traverse the right subtree 
    int y = countAP(root->right, arr); 
 
    return x + y; 
} 
 
// Driver Code 
int main() 
{ 
    Node* root = new Node(1); 
    root->left = new Node(3); 
    root->right = new Node(6); 
    root->left->left = new Node(5); 
    root->left->right = new Node(7); 
    root->right->left = new Node(11); 
    root->right->right = new Node(23); 
 
    cout << countAP(root, {}); 
 
    return 0; 
} 

Java

// Java implementation to count 
// the path which forms an A.P. 
import java.util.*;
 
class GFG{ 
 
int count = 0; 
 
// Node structure 
static class Node 
{ 
    int val; 
     
    // left and right child of the node 
    Node left, right; 
     
    // Initialization constructor 
    Node(int x) 
    { 
        val = x; 
        left = null; 
        right = null; 
    } 
}; 
 
// Function to check if path 
// format A.P. or not 
static boolean check(Vector<Integer> arr) 
{ 
    if (arr.size() == 1) 
        return true; 
 
    // If size of arr is greater than 2 
    int d = arr.get(1) - arr.get(0);
 
    for(int i = 2; i < arr.size(); i++) 
    { 
        if (arr.get(i) - arr.get(i - 1) != d) 
            return false; 
    } 
    return true; 
} 
 
// Function to find the maximum 
// setbits sum from root to leaf 
static int countAP(Node root, 
                   Vector<Integer> arr) 
{ 
    if (root == null) 
        return 0;
 
    arr.add(root.val); 
 
    // If the node is a leaf node 
    if (root.left == null && 
        root.right == null) 
    { 
        if (check(arr)) 
            return 1; 
        return 0;
    } 
 
    // Traverse left subtree 
    int x = countAP(root.left, arr); 
 
    // Traverse the right subtree 
    int y = countAP(root.right, arr); 
 
    return x + y; 
} 
 
// Driver Code 
public static void main(String[] args) 
{ 
    Node root = new Node(1); 
    root.left = new Node(3); 
    root.right = new Node(6); 
    root.left.left = new Node(5); 
    root.left.right = new Node(7); 
    root.right.left = new Node(11); 
    root.right.right = new Node(23); 
 
    System.out.print(countAP(root, new Vector<Integer>())); 
} 
} 
 
// This code is contributed by gauravrajput1

Python3

# Python3 implementation to count 
# the path which forms an A.P.
 
# Node structure
class Node:
    def __init__(self, x):
         
        self.val = x
        self.left = None
        self.right = None
         
# Function to check if path 
# format A.P. or not 
def check(arr):
     
    if len(arr) == 1:
        return True
     
    # If size of arr is greater than 2 
    d = arr[1] - arr[0]
     
    for i in range(2, len(arr)):
        if arr[i] - arr[i - 1] != d:
            return False
             
    return True
 
# Function to find the maximum 
# setbits sum from root to leaf 
def countAP(root, arr):
     
    if not root:
        return 0
     
    arr.append(root.val)
     
    # If the node is a leaf node
    if (root.left == None and
        root.right == None):
        if check(arr):
            return 1
        return 0
     
    # Traverse the left subtree
    x = countAP(root.left, arr)
     
    # Traverse the right subtree
    y = countAP(root.right, arr)
     
    return x + y
 
# Driver code
root = Node(1)
root.left = Node(3)
root.right = Node(6)
root.left.left = Node(5)
root.left.right = Node(7)
root.right.left = Node(11)
root.right.right = Node(23)
 
print(countAP(root, []))
 
# This code is contributed by stutipathak31jan

C#

// C# implementation to count 
// the path which forms an A.P. 
using System;
using System.Collections.Generic;
 
class GFG{ 
 
//int count = 0; 
 
// Node structure 
class Node 
{ 
    public int val; 
     
    // left and right child of the node 
    public Node left, right; 
     
    // Initialization constructor 
    public Node(int x) 
    { 
        val = x; 
        left = null; 
        right = null; 
    } 
}; 
 
// Function to check if path 
// format A.P. or not 
static bool check(List<int> arr) 
{ 
    if (arr.Count == 1) 
        return true; 
 
    // If size of arr is greater than 2 
    int d = arr[1] - arr[0];
 
    for(int i = 2; i < arr.Count; i++) 
    { 
        if (arr[i] - arr[i - 1] != d) 
            return false; 
    } 
    return true; 
} 
 
// Function to find the maximum 
// setbits sum from root to leaf 
static int countAP(Node root, 
                   List<int> arr) 
{ 
    if (root == null) 
        return 0;
 
    arr.Add(root.val); 
 
    // If the node is a leaf node 
    if (root.left == null && 
       root.right == null) 
    { 
        if (check(arr)) 
            return 1; 
        return 0;
    } 
 
    // Traverse left subtree 
    int x = countAP(root.left, arr); 
 
    // Traverse the right subtree 
    int y = countAP(root.right, arr); 
 
    return x + y; 
} 
 
// Driver Code 
public static void Main(String[] args) 
{ 
    Node root = new Node(1); 
    root.left = new Node(3); 
    root.right = new Node(6); 
    root.left.left = new Node(5); 
    root.left.right = new Node(7); 
    root.right.left = new Node(11); 
    root.right.right = new Node(23); 
 
    Console.Write(countAP(root, new List<int>())); 
} 
} 
 
// This code is contributed by amal kumar choubey 

Javascript

<script>
 
// JavaScript implementation to count 
// the path which forms an A.P. 
let count = 0;
 
// Node structure
class Node 
{
     
    // Initialize constructor
    constructor(x)
    {
        this.val = x;
        this.left = null;
        this.right = null;
    }
}
 
var root;
 
// Function to check if path 
// format A.P. or not 
function check(arr) 
{ 
    if (arr.length == 1) 
        return true; 
 
    // If size of arr is greater than 2 
    let d = arr[1] - arr[0]; 
 
    for(let i = 2; i < arr.length; i++)
    { 
        if (arr[i] - arr[i - 1] != d) 
            return false; 
    } 
    return true; 
} 
 
// Function to find the maximum 
// setbits sum from root to leaf 
function countAP(root, arr) 
{ 
    if (!root) 
        return 0; 
 
    arr.push(root.val); 
 
    // If the node is a leaf node 
    if (root.left == null && 
       root.right == null) 
    { 
        if (check(arr)) 
            return 1; 
        return 0; 
    } 
 
    // Traverse left subtree 
    let x = countAP(root.left, arr); 
 
    // Traverse the right subtree 
    let y = countAP(root.right, arr); 
 
    return x + y; 
} 
 
// Driver Code
root = new Node(1); 
root.left = new Node(3); 
root.right = new Node(6); 
root.left.left = new Node(5); 
root.left.right = new Node(7); 
root.right.left = new Node(11); 
root.right.right = new Node(23); 
 
let arr = [];
document.write(countAP(root, arr));
 
// This code is contributed by Dharanendra L V.
 
</script>

Output:

Time Complexity: O(N)
Auxiliary Space: O(h), where h is the height of binary tree.

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