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:

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


Output:

3


My Personal Notes arrow_drop_up

Data science |Machine learning|Programming facebook -https//wwwfacebookcom/profilephpid=100002787011326

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.




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.