Count the nodes in the given tree whose sum of digits of weight is odd

Given a tree, and the weights of all the nodes, the task is to count the number of nodes whose sum of digits of weights is odd.

**Examples:**

Input:Output:3

Node 1: digitSum(144) = 1 + 4 + 4 = 9

Node 2: digitSum(1234) = 1 + 2 + 3 + 4 = 10

Node 3: digitSum(21) = 2 + 1 = 3

Node 4: digitSum(5) = 5

Node 5: digitSum(77) = 7 + 7 = 14

Only the sum of digits of the weights of nodes 1, 3 and 4 are odd.

**Approach:** Perform dfs on the tree and for every node, check if the sum of the digits of its weight is odd. If yes then increment the 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 return the` `// sum of the digits of n` `int` `digitSum(` `int` `n)` `{` ` ` `int` `sum = 0;` ` ` `while` `(n) {` ` ` `sum += n % 10;` ` ` `n = n / 10;` ` ` `}` ` ` `return` `sum;` `}` ` ` `// Function to perform dfs` `void` `dfs(` `int` `node, ` `int` `parent)` `{` ` ` `// If sum of the digits of current node's` ` ` `// weight is odd then increment ans` ` ` `int` `sum = digitSum(weight[node]);` ` ` `if` `(sum % 2 == 1)` ` ` `ans += 1;` ` ` ` ` `for` `(` `int` `to : graph[node]) {` ` ` `if` `(to == parent)` ` ` `continue` `;` ` ` `dfs(to, node);` ` ` `}` `}` ` ` `// Driver code` `int` `main()` `{` ` ` ` ` `// Weights of the node` ` ` `weight[1] = 144;` ` ` `weight[2] = 1234;` ` ` `weight[3] = 21;` ` ` `weight[4] = 5;` ` ` `weight[5] = 77;` ` ` ` ` `// 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` `;` ` ` ` ` `static` `Vector<Integer>[] graph = ` `new` `Vector[` `100` `];` ` ` `static` `Integer[] weight = ` `new` `Integer[` `100` `];` ` ` ` ` `// Function to return the` ` ` `// sum of the digits of n` ` ` `static` `int` `digitSum(` `int` `n) ` ` ` `{` ` ` `int` `sum = ` `0` `;` ` ` `while` `(n > ` `0` `) ` ` ` `{` ` ` `sum += n % ` `10` `;` ` ` `n = n / ` `10` `;` ` ` `}` ` ` `return` `sum;` ` ` `}` ` ` ` ` `// Function to perform dfs` ` ` `static` `void` `dfs(` `int` `node, ` `int` `parent)` ` ` `{` ` ` ` ` `// If sum of the digits of current node's` ` ` `// weight is odd then increment ans` ` ` `int` `sum = digitSum(weight[node]);` ` ` `if` `(sum % ` `2` `== ` `1` `)` ` ` `ans += ` `1` `;` ` ` ` ` `for` `(` `int` `to : 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` `Vector<Integer>();` ` ` ` ` `// Weights of the node` ` ` `weight[` `1` `] = ` `144` `;` ` ` `weight[` `2` `] = ` `1234` `;` ` ` `weight[` `3` `] = ` `21` `;` ` ` `weight[` `4` `] = ` `5` `;` ` ` `weight[` `5` `] = ` `77` `;` ` ` ` ` `// 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.print(ans);` ` ` `}` `}` ` ` `// This code is contributed by Rajput-Ji` |

## Python3

`# Python3 implementation of the approach ` `ans ` `=` `0` ` ` `graph ` `=` `[[] ` `for` `i ` `in` `range` `(` `100` `)]` `weight ` `=` `[` `0` `] ` `*` `100` ` ` `# Function to return the ` `# sum of the digits of n ` `def` `digitSum(n):` ` ` `sum` `=` `0` ` ` `while` `(n):` ` ` `sum` `+` `=` `n ` `%` `10` ` ` `n ` `=` `n ` `/` `/` `10` ` ` `return` `sum` ` ` `# Function to perform dfs ` `def` `dfs(node, parent):` ` ` `global` `ans` ` ` ` ` `# If sum of the digits of current node's ` ` ` `# weight is odd then increment ans ` ` ` `sum` `=` `digitSum(weight[node]) ` ` ` `if` `(` `sum` `%` `2` `=` `=` `1` `):` ` ` `ans ` `+` `=` `1` ` ` ` ` `for` `to ` `in` `graph[node]:` ` ` `if` `(to ` `=` `=` `parent):` ` ` `continue` ` ` `dfs(to, node)` ` ` `# Driver code ` ` ` `# Weights of the node ` `weight[` `1` `] ` `=` `144` `weight[` `2` `] ` `=` `1234` `weight[` `3` `] ` `=` `21` `weight[` `4` `] ` `=` `5` `weight[` `5` `] ` `=` `77` ` ` `# 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 return the` ` ` `// sum of the digits of n` ` ` `static` `int` `digitSum(` `int` `n) ` ` ` `{` ` ` `int` `sum = 0;` ` ` `while` `(n > 0) ` ` ` `{` ` ` `sum += n % 10;` ` ` `n = n / 10;` ` ` `}` ` ` `return` `sum;` ` ` `}` ` ` ` ` `// Function to perform dfs` ` ` `static` `void` `dfs(` `int` `node, ` `int` `parent)` ` ` `{` ` ` ` ` `// If sum of the digits of current node's` ` ` `// weight is odd then increment ans` ` ` `int` `sum = digitSum(weight[node]);` ` ` `if` `(sum % 2 == 1)` ` ` `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] = 144;` ` ` `weight[2] = 1234;` ` ` `weight[3] = 21;` ` ` `weight[4] = 5;` ` ` `weight[5] = 77;` ` ` ` ` `// Edges of the tree` ` ` `graph[1].Add(2);` ` ` `graph[2].Add(3);` ` ` `graph[2].Add(4);` ` ` `graph[1].Add(5);` ` ` ` ` `dfs(1, 1);` ` ` ` ` `Console.Write(ans);` ` ` `}` `}` ` ` `// This code is contributed by PrinciRaj1992` |

**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) for 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.