Count the nodes in the given tree whose weight is even

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++

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


Output:

3

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.




My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : SHUBHAMSINGH10, Rajput-Ji

Article Tags :
Practice Tags :


Be the First to upvote.


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.