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Find the root of the sub-tree whose weighted sum is minimum
  • Last Updated : 14 Jun, 2020

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 for parent 3 = -1
Weight of sub-tree for parent 4 = 3
Weight of sub-tree for parent 5 = -2
Node 5 gives the minimum sub-tree weighted sum.

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 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, mini = INT_MAX;
  
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 minimum sub-tree sum
void findMin(int n)
{
  
    // For every node
    for (int i = 1; i <= n; i++) {
  
        // If current node's weight
        // is minimum so far
        if (mini > weight[i]) {
            mini = weight[i];
            ans = i;
        }
    }
}
  
// Driver code
int main()
{
    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);
    findMin(n);
  
    cout << ans;
  
    return 0;
}

Java




// Java implementation of the approach 
import java.util.*; 
  
class GFG 
    static int ans = 0, mini = Integer.MAX_VALUE; 
      
    @SuppressWarnings("unchecked")
    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 minimum sub-tree sum  x 
    static void findMin(int n) 
    
  
        // For every node 
        for (int i = 1; i <= n; i++) 
        
  
            // If current node's weight  x 
            // is minimum so far 
            if (mini > weight[i]) 
            
                mini = weight[i]; 
                ans = i; 
            
        
    
  
    // Driver code 
    public static void main(String[] args) 
    
          
        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); 
        findMin(n); 
  
        System.out.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>[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 minimum sub-tree sum  x 
    static void findMin(int n) 
    
   
        // For every node 
        for (int i = 1; i <= n; i++) 
        
   
            // If current node's weight  x 
            // is minimum so far 
            if (mini > weight[i]) 
            
                mini = weight[i]; 
                ans = i; 
            
        
    
   
    // Driver code 
    public static void Main(String[] args) 
    
           
        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); 
        findMin(n); 
   
        Console.Write(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 mini, graph, weight, 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 minimum sub-tree sum
def findMin(n):
    global mini, graph, weight, ans 
      
    # For every node
    for i in range(1, n + 1):
          
        # If current node's weight
        # is minimum so far
        if (mini > weight[i]):
            mini = weight[i]
            ans = i
  
# Driver code
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)
findMin(n)
  
print(ans)
  
# This code is contributed by SHUBHAMSINGH10
Output:
5

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.

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