Find the root of the sub-tree whose weighted sum XOR with X is maximum Last Updated : 11 Jul, 2025 Comments Improve Suggest changes Like Article Like Report Given a tree, and the weights of all the nodes, the task is to find the root of the sub-tree whose weighted sum XOR with given integer X is maximum.Examples: Input: X = 15 Output: 4 Weight of sub-tree for parent 1 = ((-1) + (5) + (-2) + (-1) + (3)) XOR 15 = 4 XOR 15 = 11 Weight of sub-tree for parent 2 = ((5) + (-1) + (3)) XOR 15 = 7 XOR 15 = 8 Weight of sub-tree for parent 3 = -1 XOR 15 = -16 Weight of sub-tree for parent 4 = 3 XOR 15 = 12 Weight of sub-tree for parent 5 = -2 XOR 15 = -15 Node 4 gives the maximum sub-tree weighted sum XOR X. Approach: Perform dfs on the tree, and for every node calculate the sub-tree weighted sum rooted at the current node then find the maximum (sum XOR X) value for a node.Below is the implementation of the above approach: C++ // C++ implementation of the approach #include <bits/stdc++.h> using namespace std; int ans = 0, maxi = INT_MIN; vector<int> graph[100]; vector<int> weight(100); // Function to perform dfs and update the tree // such that every node's weight is the sum of // the weights of all the nodes in the sub-tree // of the current node including itself void dfs(int node, int parent) { for (int to : graph[node]) { if (to == parent) continue; dfs(to, node); // Calculating the weighted // sum of the subtree weight[node] += weight[to]; } } // Function to find the node // having maximum sub-tree sum XOR x void findMaxX(int n, int x) { // For every node for (int i = 1; i <= n; i++) { // If current node's weight XOR x // is maximum so far if (maxi < (weight[i] ^ x)) { maxi = (weight[i] ^ x); ans = i; } } } // Driver code int main() { int x = 15; int n = 5; // Weights of the node weight[1] = -1; weight[2] = 5; weight[3] = -1; weight[4] = 3; weight[5] = -2; // Edges of the tree graph[1].push_back(2); graph[2].push_back(3); graph[2].push_back(4); graph[1].push_back(5); dfs(1, 1); findMaxX(n, x); cout << ans; return 0; } Java // Java implementation of the approach import java.util.*; class GFG { static int ans = 0, maxi = Integer.MIN_VALUE; static Vector<Integer>[] graph = new Vector[100]; static Integer[] weight = new Integer[100]; // Function to perform dfs and update the tree // such that every node's weight is the sum of // the weights of all the nodes in the sub-tree // of the current node including itself static void dfs(int node, int parent) { for (int to : graph[node]) { if (to == parent) continue; dfs(to, node); // Calculating the weighted // sum of the subtree weight[node] += weight[to]; } } // Function to find the node // having maximum sub-tree sum XOR x static void findMaxX(int n, int x) { // For every node for (int i = 1; i <= n; i++) { // If current node's weight XOR x // is maximum so far if (maxi < (weight[i] ^ x)) { maxi = (weight[i] ^ x); ans = i; } } } // Driver code public static void main(String[] args) { int x = 15; int n = 5; for (int i = 0; i < 100; i++) graph[i] = new Vector<Integer>(); // Weights of the node weight[1] = -1; weight[2] = 5; weight[3] = -1; weight[4] = 3; weight[5] = -2; // Edges of the tree graph[1].add(2); graph[2].add(3); graph[2].add(4); graph[1].add(5); dfs(1, 1); findMaxX(n, x); System.out.print(ans); } } // This code is contributed by Rajput-Ji Python # Python implementation of the approach from sys import maxsize # Function to perform dfs and update the tree # such that every node's weight is the sum of # the weights of all the nodes in the sub-tree # of the current node including itself def dfs(node, parent): global maxi, graph, weight, x, ans for to in graph[node]: if (to == parent): continue dfs(to, node) # Calculating the weighted # sum of the subtree weight[node] += weight[to] # Function to find the node # having maximum sub-tree sum XOR x def findMaxX(n, x): global maxi, graph, weight, ans # For every node for i in range(1, n + 1): # If current node's weight XOR x # is maximum so far if (maxi < (weight[i] ^ x)): maxi = (weight[i] ^ x) ans = i # Driver code ans = 0 maxi = -maxsize graph = [[] for i in range(100)] weight = [0]*100 x = 15 n = 5 # Weights of the node weight[1] = -1 weight[2] = 5 weight[3] = -1 weight[4] = 3 weight[5] = -2 # Edges of the tree graph[1].append(2) graph[2].append(3) graph[2].append(4) graph[1].append(5) dfs(1, 1) findMaxX(n, x) print(ans) # This code is contributed by SHUBHAMSINGH10 C# // C# implementation of the approach using System; using System.Collections.Generic; class GFG { static int ans = 0, maxi = int.MinValue; static List<int>[] graph = new List<int>[100]; static int[] weight = new int[100]; // Function to perform dfs and update the tree // such that every node's weight is the sum of // the weights of all the nodes in the sub-tree // of the current node including itself static void dfs(int node, int parent) { foreach (int to in graph[node]) { if (to == parent) continue; dfs(to, node); // Calculating the weighted // sum of the subtree weight[node] += weight[to]; } } // Function to find the node // having maximum sub-tree sum XOR x static void findMaxX(int n, int x) { // For every node for (int i = 1; i <= n; i++) { // If current node's weight XOR x // is maximum so far if (maxi < (weight[i] ^ x)) { maxi = (weight[i] ^ x); ans = i; } } } // Driver code public static void Main(String[] args) { int x = 15; int n = 5; for (int i = 0; i < 100; i++) graph[i] = new List<int>(); // Weights of the node weight[1] = -1; weight[2] = 5; weight[3] = -1; weight[4] = 3; weight[5] = -2; // Edges of the tree graph[1].Add(2); graph[2].Add(3); graph[2].Add(4); graph[1].Add(5); dfs(1, 1); findMaxX(n, x); Console.Write(ans); } } // This code is contributed by PrinciRaj1992 JavaScript <script> // Javascript implementation of the approach let maxi = Number.MIN_VALUE, x, ans; let graph = new Array(100); let weight = new Array(100); for(let i = 0; i < 100; i++) { graph[i] = []; weight[i] = 0; } // Function to perform dfs and update the tree // such that every node's weight is the sum of // the weights of all the nodes in the sub-tree // of the current node including itself function dfs(node, parent) { for(let to in graph[node]) { if (to == parent) continue; dfs(to, node); // Calculating the weighted // sum of the subtree weight[node] += weight[to]; } } // Function to find the node // having maximum sub-tree sum XOR x function findMaxX(n, x) { // For every node for (let i = 1; i <= n; i++) { // If current node's weight XOR x // is maximum so far if (maxi < (weight[i] ^ x)) { maxi = (weight[i] ^ x); ans = i; } } } // Driver Code x = 15; let n = 5; // Weights of the node weight[1] = -1; weight[2] = 5; weight[3] = -1; weight[4] = 3; weight[5] = -2; // Edges of the tree graph[1].push(2); graph[2].push(3); graph[2].push(4); graph[1].push(5); dfs(1, 1); findMaxX(n, x); document.write(ans); // This code is contributed by unknown2108 </script> Output: 4 Complexity Analysis: Time Complexity : O(N). In dfs, every node of the tree is processed once and hence the complexity due to the dfs is O(N) if there are total N nodes in the tree. Therefore, the time complexity is O(N).Auxiliary Space : O(n). Recursion Stack. Comment More infoAdvertise with us Next Article Find the root of the sub-tree whose weighted sum XOR with X is minimum M mohit kumar 29 Follow Improve Article Tags : DSA Bitwise-XOR DFS Practice Tags : DFS Similar Reads Find the root of the sub-tree whose weighted sum XOR with X is minimum Given a tree, and the weights of all the nodes, the task is to find the root of the sub-tree whose weighted sum XOR with given integer X is minimum.Examples: Input: X = 15 Output: 5 Weight of sub-tree for parent 1 = ((-1) + (5) + (-2) + (-1) + (3)) XOR 15 = 4 XOR 15 = 11 Weight of sub-tree for paren 7 min read Find the root of the sub-tree whose weighted sum XOR with X is minimum Given a tree, and the weights of all the nodes, the task is to find the root of the sub-tree whose weighted sum XOR with given integer X is minimum.Examples: Input: X = 15 Output: 5 Weight of sub-tree for parent 1 = ((-1) + (5) + (-2) + (-1) + (3)) XOR 15 = 4 XOR 15 = 11 Weight of sub-tree for paren 7 min read Find the root of the sub-tree whose weighted sum XOR with X is minimum Given a tree, and the weights of all the nodes, the task is to find the root of the sub-tree whose weighted sum XOR with given integer X is minimum.Examples: Input: X = 15 Output: 5 Weight of sub-tree for parent 1 = ((-1) + (5) + (-2) + (-1) + (3)) XOR 15 = 4 XOR 15 = 11 Weight of sub-tree for paren 7 min read Find the root of the sub-tree whose weighted sum is minimum Given a tree, and the weights of all the nodes, the task is to find the root of the sub-tree whose weighted sum is minimum. Examples: Input: Output: 5 Weight of sub-tree for parent 1 = ((-1) + (5) + (-2) + (-1) + (3)) = 4 Weight of sub-tree for parent 2 = ((5) + (-1) + (3)) = 7 Weight of sub-tree fo 7 min read Find the root of the sub-tree whose weighted sum is minimum Given a tree, and the weights of all the nodes, the task is to find the root of the sub-tree whose weighted sum is minimum. Examples: Input: Output: 5 Weight of sub-tree for parent 1 = ((-1) + (5) + (-2) + (-1) + (3)) = 4 Weight of sub-tree for parent 2 = ((5) + (-1) + (3)) = 7 Weight of sub-tree fo 7 min read Find the root of the sub-tree whose weighted sum is minimum Given a tree, and the weights of all the nodes, the task is to find the root of the sub-tree whose weighted sum is minimum. Examples: Input: Output: 5 Weight of sub-tree for parent 1 = ((-1) + (5) + (-2) + (-1) + (3)) = 4 Weight of sub-tree for parent 2 = ((5) + (-1) + (3)) = 7 Weight of sub-tree fo 7 min read Like