# Find the root of the sub-tree whose weighted sum XOR with X is minimum

• Last Updated : 21 Apr, 2021

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:

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students. X = 15
Output:
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 3 gives the minimum 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 minimum (sum XOR X) value for a node.
Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `int` `ans = 0, mini = INT_MAX;` `vector<``int``> graph;``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 minimum sub-tree sum XOR x``void` `findMinX(``int` `n, ``int` `x)``{` `    ``// For every node``    ``for` `(``int` `i = 1; i <= n; i++) {` `        ``// If current node's weight XOR x``        ``// is minimum so far``        ``if` `(mini > (weight[i] ^ x)) {``            ``mini = (weight[i] ^ x);``            ``ans = i;``        ``}``    ``}``}` `// Driver code``int` `main()``{``    ``int` `x = 15;``    ``int` `n = 5;` `    ``// Weights of the node``    ``weight = -1;``    ``weight = 5;``    ``weight = -1;``    ``weight = 3;``    ``weight = -2;` `    ``// Edges of the tree``    ``graph.push_back(2);``    ``graph.push_back(3);``    ``graph.push_back(4);``    ``graph.push_back(5);` `    ``dfs(1, 1);``    ``findMinX(n, x);` `    ``cout << ans;` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``import` `java.util.*;` `class` `GFG``{``    ``static` `int` `ans = ``0``, mini = Integer.MAX_VALUE;` `    ``static` `Vector[] 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 minimum sub-tree sum XOR x``    ``static` `void` `findMinX(``int` `n, ``int` `x)``    ``{` `        ``// For every node``        ``for` `(``int` `i = ``1``; i <= n; i++)``        ``{` `            ``// If current node's weight XOR x``            ``// is minimum so far``            ``if` `(mini > (weight[i] ^ x))``            ``{``                ``mini = (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();``        ` `        ``// 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``);``        ``findMinX(n, x);` `        ``System.out.print(ans);``    ``}``}` `// This code is contributed by Rajput-Ji`

## Python3

 `# Python3 implementation of the approach``ans ``=` `0``mini ``=` `2``*``*``32` `graph ``=` `[[] ``for` `i ``in` `range``(``100``)]``weight ``=` `[``0``] ``*` `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``def` `dfs(node, parent):``    ``global` `ans, mini, graph, weight, x``    ``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 minimum sub-tree sum XOR x``def` `findMinX(n, x):``    ``global` `ans, mini,graph,weight``    ` `    ``# For every node``    ``for` `i ``in` `range``(``1``, n ``+` `1``):``        ` `        ``# If current node's weight XOR x``        ``# is minimum so far``        ``if` `(mini > (weight[i] ^ x)):``            ``mini ``=` `(weight[i] ^ x)``            ``ans ``=` `i``            ` `# Driver code``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``)``findMinX(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, mini = ``int``.MaxValue;` `    ``static` `List<``int``>[] graph = ``new` `List<``int``>;``    ``static` `int``[] weight = ``new` `int``;` `    ``// 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 minimum sub-tree sum XOR x``    ``static` `void` `findMinX(``int` `n, ``int` `x)``    ``{` `        ``// For every node``        ``for` `(``int` `i = 1; i <= n; i++)``        ``{` `            ``// If current node's weight XOR x``            ``// is minimum so far``            ``if` `(mini > (weight[i] ^ x))``            ``{``                ``mini = (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;``        ``weight = 5;``        ``weight = -1;``        ``weight = 3;``        ``weight = -2;` `        ``// Edges of the tree``        ``graph.Add(2);``        ``graph.Add(3);``        ``graph.Add(4);``        ``graph.Add(5);` `        ``dfs(1, 1);``        ``findMinX(n, x);` `        ``Console.Write(ans);``    ``}``}` `// This code is contributed by PrinciRaj1992`

## Javascript

 ``
Output:
`3`

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 space can be up to O(n).

My Personal Notes arrow_drop_up