Cuente los Nodes en el árbol dado cuyo peso es una potencia de dos

Dado un árbol y los pesos de todos los Nodes, la tarea es contar el número de Nodes cuyo peso es una potencia de 2.
Ejemplos: 
 

Aporte: 
 

Salida: 1 
Solo el peso del Node 4 es potencia de 2. 
 

Enfoque: Realice dfs en el árbol y para cada Node, verifique si su peso es una potencia de 2 o no, en caso afirmativo, 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 to perform dfs
void dfs(int node, int parent)
{
    // If weight of the current node
    // is a power of 2
    int x = weight[node];
    if (x && (!(x & (x - 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] = 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;
 
    @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 a power of 2
        int x = weight[node];
        if (x != 0 && (x & (x - 1)) == 0)
            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<>();
 
        // 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
// sanjeev2552

// 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 to perform dfs
static void dfs(int node, int parent)
{
 
    // If weight of the current node
    // is a power of 2
    int x = weight[node];
    bool result = Convert.ToBoolean((x & (x - 1)));
    bool result1 = Convert.ToBoolean(x);
    if (result1 && (!result))
        ans += 1;
 
    for (int i = 0; i < graph[node].Count; i++)
    {
        if (graph[node][i] == parent)
            continue;
        dfs(graph[node][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 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 shubhamsingh10

# 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 mini, graph, weight, ans
     
    # If weight of the current node
    # is a power of 2
    x = weight[node]
    if (x and (not (x & (x - 1)))):
        ans += 1
    for to in graph[node]:
        if (to == parent):
            continue
        dfs(to, node)
         
        # Calculating the weighted
        # sum of the subtree
        weight[node] += weight[to]
     
# 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

<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 a power of 2
    let x = weight[node];
    if (x && (!(x & (x - 1))))
        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>

1

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.