Count the number of paths from root to leaf of a Binary tree with given XOR value

Given a value K and a binary tree, we have to find out the total number of paths from root to leaf nodes having XOR of all its nodes along the path equal to K.

Examples:

Input: K = 6
       2
      / \
     1   4
    / \   
   10  5   
Output: 2
Explanation:
Subtree 1: 
   2
    \
     4
This particular path has 2 nodes, 2 and 4 
and (2 xor 4) = 6.

Subtree 2:
        2
       /
      1
       \
        5
This particular path has 3 nodes; 2, 1 and 5 
and (2 xor 1 xor 5) = 6.

Approach:

To solve the question mentioned above we have to traverse the tree recursively using pre-order traversal. For each node keep calculating the XOR of the path from root till the current node.

XOR of current node’s path = (XOR of the path till the parent) ^ (current node value)



If the node is a leaf node that is left and the right child for the current nodes are NULL then we check if the xor value of path is K, if it is then we increase the count otherwise we do nothing. Finally, print the value in the count.

Below is the implementation of the above approach:

C++

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// C++ program to Count the number of
// path from the root to leaf of a
// Binary tree with given XOR value
  
#include <bits/stdc++.h>
using namespace std;
  
// Binary tree node
struct Node {
    int data;
  
    struct Node *left, *right;
};
  
// Function to create a new node
struct Node* newNode(int data)
{
    struct Node* newNode = new Node;
  
    newNode->data = data;
  
    newNode->left
        = newNode->right = NULL;
  
    return (newNode);
}
  
void Count(Node* root, int xr, int& res, int& k)
{
  
    // updating the xor value
    // with the xor of the path from
    // root to the node
    xr = xr ^ root->data;
  
    // check if node is leaf node
    if (root->left == NULL && root->right == NULL) {
  
        if (xr == k) {
            res++;
        }
        return;
    }
  
    // check if the left
    // node exist in the tree
    if (root->left != NULL) {
        Count(root->left, xr, res, k);
    }
  
    // check if the right node
    // exist in the tree
    if (root->right != NULL) {
        Count(root->right, xr, res, k);
    }
  
    return;
}
  
// Function to find the required count
int findCount(Node* root, int K)
{
  
    int res = 0, xr = 0;
  
    // recursively traverse the tree
    // and compute the count
    Count(root, xr, res, K);
  
    // return the result
    return res;
}
  
// Driver code
int main(void)
{
    // Create the binary tree
    struct Node* root = newNode(2);
    root->left = newNode(1);
    root->right = newNode(4);
    root->left->left = newNode(10);
    root->left->right = newNode(5);
  
    int K = 6;
  
    cout << findCount(root, K);
  
    return 0;
}

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Java

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// Java program to Count the number of
// path from the root to leaf of a
// Binary tree with given XOR value
import java.util.*;
  
class GFG{
   
// Binary tree node
static class Node {
    int data;
   
    Node left, right;
};
static int res, k; 
  
// Function to create a new node
static Node newNode(int data)
{
    Node newNode = new Node();
   
    newNode.data = data;
   
    newNode.left
        = newNode.right = null;
   
    return (newNode);
}
   
static void Count(Node root, int xr)
{
   
    // updating the xor value
    // with the xor of the path from
    // root to the node
    xr = xr ^ root.data;
   
    // check if node is leaf node
    if (root.left == null && root.right == null) {
   
        if (xr == k) {
            res++;
        }
        return;
    }
   
    // check if the left
    // node exist in the tree
    if (root.left != null) {
        Count(root.left, xr);
    }
   
    // check if the right node
    // exist in the tree
    if (root.right != null) {
        Count(root.right, xr);
    }
   
    return;
}
   
// Function to find the required count
static int findCount(Node root, int K)
{
   
    int xr = 0;
    res = 0;
    k = K;
  
    // recursively traverse the tree
    // and compute the count
    Count(root, xr);
   
    // return the result
    return res;
}
   
// Driver code
public static void main(String[] args)
{
    // Create the binary tree
    Node root = newNode(2);
    root.left = newNode(1);
    root.right = newNode(4);
    root.left.left = newNode(10);
    root.left.right = newNode(5);
   
    int K = 6;
   
    System.out.print(findCount(root, K));
}
}
  
// This code is contributed by 29AjayKumar

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C#

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// C# program to Count the number of
// path from the root to leaf of a
// Binary tree with given XOR value
using System;
  
class GFG{
    
// Binary tree node
class Node {
    public int data;
    
    public Node left, right;
};
static int res, k; 
   
// Function to create a new node
static Node newNode(int data)
{
    Node newNode = new Node();
    
    newNode.data = data;
    
    newNode.left
        = newNode.right = null;
    
    return (newNode);
}
    
static void Count(Node root, int xr)
{
    
    // updating the xor value
    // with the xor of the path from
    // root to the node
    xr = xr ^ root.data;
    
    // check if node is leaf node
    if (root.left == null && root.right == null) {
    
        if (xr == k) {
            res++;
        }
        return;
    }
    
    // check if the left
    // node exist in the tree
    if (root.left != null) {
        Count(root.left, xr);
    }
    
    // check if the right node
    // exist in the tree
    if (root.right != null) {
        Count(root.right, xr);
    }
    
    return;
}
    
// Function to find the required count
static int findCount(Node root, int K)
{
    
    int xr = 0;
    res = 0;
    k = K;
   
    // recursively traverse the tree
    // and compute the count
    Count(root, xr);
    
    // return the result
    return res;
}
    
// Driver code
public static void Main(String[] args)
{
    // Create the binary tree
    Node root = newNode(2);
    root.left = newNode(1);
    root.right = newNode(4);
    root.left.left = newNode(10);
    root.left.right = newNode(5);
    
    int K = 6;
    
    Console.Write(findCount(root, K));
}
}
  
// This code is contributed by Princi Singh

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Output:

2

Time Complexity: As in the above approach, we are iterating over each node only once, therefore it will take O(N) time where N is the number of nodes in the Binary tree.

Auxiliary Space: As in the above approach there is no extra space used, therefore the Auxillary Space complexity will be O(1).

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