计算给定树中权重为2的幂的节点

给定一棵树，以及所有节点的权重，任务是计算权重为2的幂的节点数。

例子：

Input:

Output: 1
Only the weight of the node 4 is a power of 2.

方法：在树上执行dfs，对于每个节点，检查其权重是否为2的幂，如果是，则增加计数。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
  
int ans = 0;
  
vector graph[100];
vector 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

// Java implementation of the approach 
import java.util.*;
  
class GFG 
{
  
    static int ans = 0;
  
    @SuppressWarnings("unchecked")
    static Vector[] 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#

// C# implementation of the approach 
using System; 
using System.Collections.Generic; 
  
class GFG 
{ 
      
static int ans = 0; 
static List> graph = new List>(); 
static List weight = new List(); 
  
// 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()); 
  
    // 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

# 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

输出：

1

复杂度分析：

• 时间复杂度： O(N)。
  在DFS中，树的每个节点都被处理一次，因此对于树中的N个节点，由于DFS而导致的复杂度为O(N)。因此，时间复杂度为O(N)。
• 辅助空间： O(1)。
  不需要任何额外的空间，因此空间复杂度是恒定的。