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Count the nodes in the given tree whose weight is even
  • Last Updated : 12 Aug, 2020

Given a tree, and the weights of all the nodes, the task is to count the number of nodes whose weight is even.

Examples:

Input:

Output: 3
Only the weights of the nodes 2, 4 and 5 are even.

Approach: Perform dfs on the tree and for every node, check if it’s weight is divisible by 2 or not. If yes then increment count.

Below is the implementation of the above approach:



C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
int ans = 0;
  
vector<int> graph[100];
vector<int> weight(100);
  
// Function to perform dfs
void dfs(int node, int parent)
{
    // If weight of the current node is even
    if (weight[node] % 2 == 0)
        ans += 1;
  
    for (int to : graph[node]) {
        if (to == parent)
            continue;
        dfs(to, node);
    }
}
  
// Driver code
int main()
{
    int x = 15;
  
    // Weights of the node
    weight[1] = 5;
    weight[2] = 10;
    weight[3] = 11;
    weight[4] = 8;
    weight[5] = 6;
  
    // 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);
  
    cout << ans;
  
    return 0;
}

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Java

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// Java implementation of the approach 
import java.util.*; 
  
class GFG 
    static int ans = 0
      
    @SuppressWarnings("unchecked"
    static Vector<Integer>[] graph = new Vector[100]; 
    static int[] weight = new int[100];
      
    // Function to perform dfs 
    static void dfs(int node, int parent) 
    
        // If weight of the current node is even 
        if (weight[node] % 2 == 0
            ans += 1
      
        for (int to : graph[node]) 
        
            if (to == parent) 
                continue
            dfs(to, node); 
        
    
      
    // Driver code 
    public static void main(String[] args)
    
        int x = 15
      
    for (int i = 0; i < 100; i++) 
            graph[i] = new Vector<>(); 
          
        // Weights of the node 
        weight[1] = 5
        weight[2] = 10
        weight[3] = 11
        weight[4] = 8
        weight[5] = 6
      
        // Edges of the tree 
        graph[1].add(2); 
        graph[2].add(3); 
        graph[2].add(4); 
        graph[1].add(5); 
      
        dfs(1, 1); 
      
        System.out.println(ans);
    
}
  
// This code is contributed by shubhamsingh10

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Python3

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# Python3 implementation of the approach
ans = 0
  
graph = [[] for i in range(100)]
weight = [0] * 100
  
# Function to perform dfs
def dfs(node, parent):
    global ans
      
    # If weight of the current node is even
    if (weight[node] % 2 == 0):
        ans += 1
      
    for to in graph[node]:
        if (to == parent):
            continue
        dfs(to, node)
  
# Driver code
x = 15
  
# Weights of the node
weight[1] = 5
weight[2] = 10
weight[3] = 11
weight[4] = 8
weight[5] = 6
  
# Edges of the tree
graph[1].append(2)
graph[2].append(3)
graph[2].append(4)
graph[1].append(5)
  
dfs(1, 1)
print(ans)
  
# This code is contributed by SHUBHAMSINGH10

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C#

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// C# implementation of the approach 
using System;
using System.Collections.Generic;
  
class GFG 
    static int ans = 0; 
    static List<int>[] graph = new List<int>[100]; 
    static int[] weight = new int[100];
       
    // Function to perform dfs 
    static void dfs(int node, int parent) 
    
        // If weight of the current node is even 
        if (weight[node] % 2 == 0) 
            ans += 1; 
       
        foreach (int to in graph[node]) 
        
            if (to == parent) 
                continue
            dfs(to, node); 
        
    
       
    // Driver code 
    public static void Main(String[] args)
    {      
        for (int i = 0; i < 100; i++) 
            graph[i] = new List<int>(); 
           
        // Weights of the node 
        weight[1] = 5; 
        weight[2] = 10; 
        weight[3] = 11; 
        weight[4] = 8; 
        weight[5] = 6; 
       
        // Edges of the tree 
        graph[1].Add(2); 
        graph[2].Add(3); 
        graph[2].Add(4); 
        graph[1].Add(5); 
       
        dfs(1, 1); 
       
        Console.WriteLine(ans);
    
}
  
// This code is contributed by Rajput-Ji

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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(1).
    Any extra space is not required, so the space complexity is constant.

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