Maximum XOR path of a Binary Tree

Given a Binary Tree, the task is to find the maximum of all the XOR value of all the nodes in the path from the root to leaf.


      / \
     1   4
    / \   
   10  8   
Output: 11
All the paths are: 
2-1-10 XOR-VALUE = 9
2-1-8 XOR-VALUE = 11
2-4 XOR-VALUE = 6

      /   \
     1     4
    / \   / \
   10  8 5  10
Output: 12


  1. 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)

  2. If the node is a leaf node that is left and the right child for the current nodes are NULL then we compute the max-Xor, as

    max-Xor = max(max-Xor, cur-Xor).

Below is the implementation of the above approach:






// C++ program to compute the
// Max-Xor value of path from
// the root to leaf of a Binary tree
#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->right = NULL;
    return (newNode);
// Function calculate the
// value of max-xor
void Solve(Node* root, int xr,
           int& max_xor)
    // 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) {
        max_xor = max(max_xor, xr);
    // Check if the left
    // node exist in the tree
    if (root->left != NULL) {
        Solve(root->left, xr,
    // Check if the right node
    // exist in the tree
    if (root->right != NULL) {
        Solve(root->right, xr,
// Function to find the
// required count
int findMaxXor(Node* root)
    int xr = 0, max_xor = 0;
    // Recursively traverse
    // the tree and compute
    // the max_xor
    Solve(root, xr, max_xor);
    // Return the result
    return max_xor;
// 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(8);
    root->right->left = newNode(5);
    root->right->right = newNode(10);
    cout << findMaxXor(root);
    return 0;




Time Complexity: 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 Complexity: The Auxillary Space complexity will be O(1), as there is no extra space used


My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using or mail your article to See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.