# 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:

`// 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; ` `} ` |

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**Output:**

3

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