计算给定树中权重为偶校验的节点

给定一棵树和所有节点的权重，任务是计算权重为偶校验的节点的数量，即它们中设置位的计数是否为偶数。
例子：

Input: 
 

Output: 3 
 

Weight	Binary Representation	Parity
5	0101	Even
10	1010	Even
11	1011	Odd
8	1000	Odd
6	0110	Even

 

方法：在树上执行 dfs，对于每个节点，检查它的权重是否为偶校验。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 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

// 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 weight = new List();
 
// 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());
     
    // 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

Javascript

  
// 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.
      

输出：

3

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 的复杂性是 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。

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