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` `] ` `=` `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*

*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

__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.

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 the nodes in the given tree whose weight is even parity
- Count of all prime weight nodes between given nodes in the given Tree
- Count number of paths whose weight is exactly X and has at-least one edge of weight M
- Count the nodes in the given tree whose weight is prime
- Count the nodes in the given tree whose weight is a power of two
- 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 nodes in the given tree whose weight is a fibonacci number
- Count the nodes in the given tree whose weight is a powerful number
- Count the nodes in the given Tree whose weight is a Perfect Number
- Count the nodes whose weight is a perfect square
- 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.
- Print even positioned nodes of even levels in level order of the given binary tree
- Count the nodes of the tree which make a pangram when concatenated with the sub-tree nodes
- Count the nodes of the given tree whose weighted string is a palindrome
- Count the nodes of a tree whose weighted string is an anagram of the given string
- Count the nodes of the tree whose weighted string contains a vowel
- Count the nodes of a tree whose weighted string does not contain any duplicate characters
- Count of nodes in a Binary Tree whose child is its prime factors

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.