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++
// 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
filter_none
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 |
chevron_right
filter_none
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] = 5weight[2] = 10weight[3] = 11weight[4] = 8weight[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
filter_none
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 |
chevron_right
filter_none
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.
Recommended Posts:
- Count of all prime weight nodes between given nodes in the given Tree
- Count the nodes in the given tree whose sum of digits of weight is odd
- Count the nodes of the given tree whose weight has X as a factor
- Count the nodes in the given tree whose weight is a power of two
- Count the nodes in the given tree whose weight is even parity
- Count the nodes in the given tree whose weight is prime
- Count the nodes in the given Tree whose weight is a Perfect Number
- Count the nodes in the given tree whose weight is a powerful number
- Count nodes in the given tree whose weight is a fibonacci number
- Count of Nodes which has Prime Digit sum weight in a Tree
- Query to find the maximum and minimum weight between two nodes in the given tree using LCA.
- Count the nodes whose weight is a perfect square
- Count the nodes of the tree which make a pangram when concatenated with the sub-tree nodes
- Count number of paths whose weight is exactly X and has at-least one edge of weight M
- Count the number of nodes at given level in a tree using BFS.
- Count Non-Leaf nodes in a Binary Tree
- Determine the count of Leaf nodes in an N-ary tree
- Count the number of nodes at a given level in a tree using DFS
- Count of nodes in a Binary tree with immediate children as its factors
- Count the number of visible nodes in Binary Tree
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.
