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Count the nodes in the given tree whose weight is even

  • Last Updated : 22 Apr, 2021

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:
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++




// 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;
}

Java




// 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

Python3




# 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

C#




// 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

Javascript




<script>
// Javascript implementation of the approach
    let ans = 0;
    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
    function dfs(node, parent)
    {
         
        // If weight of the current node is even
        if (weight[node] % 2 == 0)
        {
            ans += 1;
        }
       
        for (let to=0;to<graph[node].length ;to++)
        {
         
            if (graph[node][to] == parent)
               {
                continue;
            }
            dfs(graph[node][to], node);
        }
    }
     
    // Driver code
    let 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(2);
        graph[2].push(3);
        graph[2].push(4);
        graph[1].push(5);
        dfs(1, 1);
        document.write(ans);
         
    // This code is contributed by unknown2108
</script>
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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