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 nodestruct Node { int data; struct Node *left, *right;};// Function to create a new nodestruct 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-xorvoid 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 countint 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 codeint 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 treeclass 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 nodeclass 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-xordef 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 treeroot = 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 approachusing 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 nodeclass Node { constructor() { this.data = 0; this.left = null; this.right = null; }};// Function to create a new nodefunction newNode(data){ var newNode = new Node; newNode.data = data; newNode.left = newNode.right = null; return (newNode);}// Function calculate the// value of Math.max-xorfunction 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 countfunction 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 treevar 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:
12
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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