Maximum XOR path of a Binary Tree

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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.
Examples:

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

Input: 
        2
      /   \
     1     4
    / \   / \
   10  8 5  10
Output: 12

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 compute the max-Xor, as

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

Below is the implementation of the above approach:

C++

// 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->left
        = 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);
        return;
    }
 
    // Check if the left
    // node exist in the tree
    if (root->left != NULL) {
        Solve(root->left, xr,
              max_xor);
    }
 
    // Check if the right node
    // exist in the tree
    if (root->right != NULL) {
        Solve(root->right, xr,
              max_xor);
    }
 
    return;
}
 
// 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;
}

Java

// Java program to compute the
// Max-Xor value of path from
// the root to leaf of a Binary tree
class GFG {
 
  // Binary tree node
  static class Node {
    int data = 0;
 
    Node left = null, right = null;
  };
 
  // 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);
  }
 
  // Function calculate the
  // value of max-xor
  static int 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 = Math.max(max_xor, xr);
      return max_xor;
    }
 
    // Check if the left
    // node exist in the tree
    if (root.left != null) {
      max_xor = Solve(root.left, xr,
                      max_xor);
    }
 
    // Check if the right node
    // exist in the tree
    if (root.right != null) {
      max_xor = Solve(root.right, xr,
                      max_xor);
    }
 
    return max_xor;
  }
 
  // Function to find the
  // required count
  static int findMaxXor(Node root) {
 
    int xr = 0, max_xor = 0;
 
    // Recursively traverse
    // the tree and compute
    // the max_xor
    max_xor = Solve(root, xr, max_xor);
 
    // Return the result
    return max_xor;
  }
 
  // 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(8);
    root.right.left = newNode(5);
    root.right.right = newNode(10);
 
    System.out.print(findMaxXor(root));
  }
}
 
// This code is contributed by saurabh_jaiswal.

Python3

# Python3 program to compute the 
# Max-Xor value of path from 
# the root to leaf of a Binary tree 
 
# Binary tree node
class Node:
     
    # Function to create a new node
    def __init__(self, data):
         
        self.data = data
        self.left = None
        self.right = None
 
# Function calculate the 
# value of max-xor
def Solve(root, xr, 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 == None and
        root.right == None):
        max_xor[0] = max(max_xor[0], xr)
     
    # Check if the left 
    # node exist in the tree
    if root.left != None:
        Solve(root.left, xr, max_xor)
     
    # Check if the right node 
    # exist in the tree 
    if root.right != None:
        Solve(root.right, xr, max_xor)
         
    return
 
# Function to find the 
# required count 
def findMaxXor(root):
     
    xr, max_xor = 0, [0]
     
    # Recursively traverse 
    # the tree and compute 
    # the max_xor 
    Solve(root, xr, max_xor)
     
    # Return the result
    return max_xor[0]
 
# Driver code
 
# Create the binary tree
root = Node(2)
root.left = Node(1)
root.right = Node(4) 
root.left.left = Node(10) 
root.left.right = Node(8) 
root.right.left = Node(5) 
root.right.right = Node(10) 
 
print(findMaxXor(root))
 
# This code is contributed by Shivam Singh

C#

// C# code for the above approach
using System;
 
namespace GFG
{
  // Binary tree node
  class Node
  {
    public int data = 0;
    public Node left = null, right = null;
  }
 
  class GFG
  {
     
    // 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);
    }
 
    // Function calculate the
    // value of max-xor
    static int 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 = Math.Max(max_xor, xr);
        return max_xor;
      }
 
      // Check if the left
      // node exist in the tree
      if (root.left != null)
      {
        max_xor = Solve(root.left, xr,
                        max_xor);
      }
 
      // Check if the right node
      // exist in the tree
      if (root.right != null)
      {
        max_xor = Solve(root.right, xr,
                        max_xor);
      }
 
      return max_xor;
    }
 
    // Function to find the
    // required count
    static int findMaxXor(Node root)
    {
      int xr = 0, max_xor = 0;
 
      // Recursively traverse
      // the tree and compute
      // the max_xor
      max_xor = Solve(root, xr, max_xor);
 
      // Return the result
      return max_xor;
    }
 
    // Driver code
    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(8);
      root.right.left = newNode(5);
      root.right.right = newNode(10);
 
      Console.WriteLine(findMaxXor(root));
    }
  }
}
 
// This code is contributed by Potta Lokesh

Javascript

<script>
  
// JavaScript program to compute the
// Max-Xor value of path from
// the root to leaf of a Binary tree
 
// Binary tree node
class Node {
 
    constructor()
    {
        this.data = 0;
        this.left = null;
        this.right = null;
    }
};
 
// Function to create a new node
function newNode(data)
{
    var newNode = new Node;
 
    newNode.data = data;
 
    newNode.left
        = newNode.right = null;
 
    return (newNode);
}
 
// Function calculate the
// value of Math.max-xor
function Solve(root, xr, 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 = Math.max(max_xor, xr);
        return max_xor;
    }
 
    // Check if the left
    // node exist in the tree
    if (root.left != null) {
        max_xor = Solve(root.left, xr,
              max_xor);
    }
 
    // Check if the right node
    // exist in the tree
    if (root.right != null) {
       max_xor = Solve(root.right, xr,
              max_xor);
    }
 
    return max_xor;
}
 
// Function to find the
// required count
function findMaxXor(root)
{
 
    var xr = 0, max_xor = 0;
 
    // Recursively traverse
    // the tree and compute
    // the max_xor
    max_xor = Solve(root, xr, max_xor);
 
    // Return the result
    return max_xor;
}
 
// Driver code
// Create the binary tree
var 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);
document.write( findMaxXor(root));
 
</script>

Output:

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 Auxiliary Space complexity will be O(1), as there is no extra space used

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