# 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.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `using` `namespace` `std; ` ` `  `int` `ans = 0; ` ` `  `vector<``int``> graph; ` `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 = 5; ` `    ``weight = 10; ` `    ``weight = 11; ` `    ``weight = 8; ` `    ``weight = 6; ` ` `  `    ``// Edges of the tree ` `    ``graph.push_back(2); ` `    ``graph.push_back(3); ` `    ``graph.push_back(4); ` `    ``graph.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[] 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``>;  ` `    ``static` `int``[] weight = ``new` `int``; ` `      `  `    ``// 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 = 5;  ` `        ``weight = 10;  ` `        ``weight = 11;  ` `        ``weight = 8;  ` `        ``weight = 6;  ` `      `  `        ``// Edges of the tree  ` `        ``graph.Add(2);  ` `        ``graph.Add(3);  ` `        ``graph.Add(4);  ` `        ``graph.Add(5);  ` `      `  `        ``dfs(1, 1);  ` `      `  `        ``Console.WriteLine(ans); ` `    ``}  ` `} ` ` `  `// This code is contributed by Rajput-Ji `

Output:

```3
```

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