# 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:

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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;` `}` |

## 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` |

## 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` `>[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` |

## Javascript

`<script>` `// Javascript implementation of the approach` ` ` `let ans = 0;` ` ` `let graph = ` `new` `Array(100);` ` ` `let weight = ` `new` `Array(100);` ` ` `for` `(let i = 0; i < 100; i++)` ` ` `{` ` ` `graph[i] = [];` ` ` `weight[i] = 0;` ` ` `}` ` ` ` ` `// Function to perform dfs` ` ` `function` `dfs(node, parent)` ` ` `{` ` ` ` ` `// If weight of the current node is even` ` ` `if` `(weight[node] % 2 == 0)` ` ` `{` ` ` `ans += 1;` ` ` `}` ` ` ` ` `for` `(let to=0;to<graph[node].length ;to++)` ` ` `{` ` ` ` ` `if` `(graph[node][to] == parent)` ` ` `{` ` ` `continue` `;` ` ` `}` ` ` `dfs(graph[node][to], node);` ` ` `}` ` ` `}` ` ` ` ` `// Driver code` ` ` `let 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(2);` ` ` `graph[2].push(3);` ` ` `graph[2].push(4);` ` ` `graph[1].push(5);` ` ` `dfs(1, 1);` ` ` `document.write(ans);` ` ` ` ` `// This code is contributed by unknown2108` `</script>` |

**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.