# Count the nodes in the given tree whose weight is even parity

• Last Updated : 11 Jun, 2021

Given a tree and the weights of all the nodes, the task is to count the number of nodes whose weights are even parity i.e. whether the count of set bits in them is even.
Examples:

Input: Output:

Approach: Perform dfs on the tree and for every node, check if it’s weight is even parity 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 that returns true if count``// of set bits in x is even``bool` `isEvenParity(``int` `x)``{``    ``// parity will store the``    ``// count of set bits``    ``int` `parity = 0;``    ``while` `(x != 0) {``        ``x = x & (x - 1);``        ``parity++;``    ``}` `    ``if` `(parity % 2 == 0)``        ``return` `true``;``    ``else``        ``return` `false``;``}` `// Function to perform dfs``void` `dfs(``int` `node, ``int` `parent)``{``    ``// If weight of the current``    ``// node has even parity``    ``if` `(isEvenParity(weight[node]))``        ``ans += 1;` `    ``for` `(``int` `to : graph[node]) {``        ``if` `(to == parent)``            ``continue``;``        ``dfs(to, node);``    ``}``}` `// Driver code``int` `main()``{``    ``// 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``;` `static` `Vector> graph = ``new` `Vector>();``static` `Vector weight = ``new` `Vector();` `// Function that returns true if count``// of set bits in x is even``static` `boolean` `isEvenParity(``int` `x)``{``    ``// parity will store the``    ``// count of set bits``    ``int` `parity = ``0``;``    ``while` `(x != ``0``)``    ``{``        ``x = x & (x - ``1``);``        ``parity++;``    ``}` `    ``if` `(parity % ``2` `== ``0``)``        ``return` `true``;``    ``else``        ``return` `false``;``}` `// Function to perform dfs``static` `void` `dfs(``int` `node, ``int` `parent)``{``    ``// If weight of the current``    ``// node has even parity``    ``if` `(isEvenParity(weight.get(node) ))``        ``ans += ``1``;` `    ``for` `(``int` `i = ``0``; i < graph.get(node).size(); i++)``    ``{``        ``if` `(graph.get(node).get(i) == parent)``            ``continue``;``        ``dfs(graph.get(node).get(i) , node);``    ``}``}` `// Driver code``public` `static` `void` `main(String args[])``{``    ``// Weights of the node``    ``weight.add( ``0``);``    ``weight.add( ``5``);``    ``weight.add( ``10``);;``    ``weight.add( ``11``);;``    ``weight.add( ``8``);``    ``weight.add( ``6``);` `    ``for``(``int` `i=``0``;i<``100``;i++)``    ``graph.add(``new` `Vector());``    ` `    ``// Edges of the tree``    ``graph.get(``1``).add(``2``);``    ``graph.get(``2``).add(``3``);``    ``graph.get(``2``).add(``4``);``    ``graph.get(``1``).add(``5``);` `    ``dfs(``1``, ``1``);` `    ``System.out.println( ans );` `}``}` `// This code is contributed by Arnab Kundu`

## Python3

 `# Python3 implementation of the approach``ans ``=` `0` `graph ``=` `[[] ``for` `i ``in` `range``(``100``)]``weight ``=` `[``0``]``*``100` `# Function that returns True if count``# of set bits in x is even``def` `isEvenParity(x):` `    ``# parity will store the``    ``# count of set bits``    ``parity ``=` `0``    ``while` `(x !``=` `0``):``        ``x ``=` `x & (x ``-` `1``)``        ``parity ``+``=` `1``        ` `    ``if` `(parity ``%` `2` `=``=` `0``):``        ``return` `True``    ``else``:``        ``return` `False` `# Function to perform dfs``def` `dfs(node, parent):``    ``global` `ans``    ` `    ``# If weight of the current``    ``# node has even parity``    ``if` `(isEvenParity(weight[node])):``        ``ans ``+``=` `1``    ` `    ``for` `to ``in` `graph[node]:``        ``if` `(to ``=``=` `parent):``            ``continue``        ``dfs(to, node)` `# Driver code` `# 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> graph = ``new` `List>();``static` `List<``int``> weight = ``new` `List<``int``>();` `// Function that returns true if count``// of set bits in x is even``static` `bool` `isEvenParity(``int` `x)``{``    ``// parity will store the``    ``// count of set bits``    ``int` `parity = 0;``    ``while` `(x != 0)``    ``{``        ``x = x & (x - 1);``        ``parity++;``    ``}` `    ``if` `(parity % 2 == 0)``        ``return` `true``;``    ``else``        ``return` `false``;``}` `// Function to perform dfs``static` `void` `dfs(``int` `node, ``int` `parent)``{``    ``// If weight of the current``    ``// node has even parity``    ``if` `(isEvenParity(weight[node]))``        ``ans += 1;` `    ``for` `(``int` `i = 0; i < graph[node].Count; i++)``    ``{``        ``if` `(graph[node][i] == parent)``            ``continue``;``        ``dfs(graph[node][i] , node);``    ``}``}` `// Driver code``static` `void` `Main()``{``    ``// Weights of the node``    ``weight.Add(0);``    ``weight.Add(5);``    ``weight.Add(10);``    ``weight.Add(11);``    ``weight.Add(8);``    ``weight.Add(6);` `    ``for``(``int` `i = 0; i < 100; i++)``    ``graph.Add(``new` `List<``int``>());``    ` `    ``// 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 mits`

## Javascript

 ``

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.

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