Cuente los Nodes en el árbol dado cuyo peso es incluso paridad

Dado un árbol y los pesos de todos los Nodes, la tarea es contar el número de Nodes cuyos pesos son pares, es decir, si el número de bits establecidos en ellos es par.
Ejemplos: 
 

Aporte: 
 

Salida: 3 
 

Peso	Representación binaria	Paridad
5	0101	Incluso
10	1010	Incluso
11	1011	Extraño
8	1000	Extraño
6	0110	Incluso

 

Enfoque: Realice dfs en el árbol y para cada Node, verifique si su peso es paritario o no. Si es así, entonces incremente el conteo.
A continuación se muestra la implementación del enfoque anterior:
 

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
int ans = 0;
 
vector<int> graph[100];
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[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 implementation of the approach
import java.util.*;
 
class GFG
{
 
static int ans = 0;
 
static Vector<Vector<Integer>> graph = new Vector<Vector<Integer>>();
static Vector<Integer> weight = new Vector<Integer>();
 
// 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<Integer>());
     
    // 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 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# implementation of the approach
using System;
using System.Collections.Generic;
 
class GFG
{
 
static int ans = 0;
 
static List<List<int>> graph = new List<List<int>>();
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[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 mits

<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 that returns true if count
// of set bits in x is even
function isEvenParity(x)
{
    // parity will store the
    // count of set bits
    let parity = 0;
    while (x != 0) {
        x = x & (x - 1);
        parity++;
    }
 
    if (parity % 2 == 0)
        return true;
    else
        return false;
}
 
// Function to perform dfs
function dfs(node, parent)
{
    // If weight of the current
    // node has even parity
    if (isEvenParity(weight[node]))
        ans += 1;
 
    for(let to=0;to<graph[node].length;to++) {
        if(graph[node][to] == parent)
            continue
        dfs(graph[node][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].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 Dharanendra L V.
      
</script>

3

Análisis de Complejidad: 
 

• Complejidad temporal: O(N). 
  En DFS, cada Node del árbol se procesa una vez y, por lo tanto, la complejidad debida al DFS es O(N) para N Nodes en el árbol. Por lo tanto, la complejidad del tiempo es O(N).
• Espacio Auxiliar: O(1). 
  No se requiere ningún espacio adicional, por lo que la complejidad del espacio es constante.