查找其与x的xor给出最大值的节点

给定一棵树，以及所有节点的权重和整数x ，任务是找到一个节点i ，以使weight [i] xor x最大。

例子：

Input:

x = 15
Output: 1
Node 1: 5 xor 15 = 10
Node 2: 10 xor 15 = 5
Node 3: 11 xor 15 = 4
Node 4: 8 xor 15 = 7
Node 5: 6 xor 15 = 9

方法：在树上执行dfs并跟踪其加权xor与x给出最大值的节点。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
  
int maximum = INT_MIN, x, ans;
  
vector graph[100];
vector weight(100);
  
// Function to perform dfs to find
// the maximum xored value
void dfs(int node, int parent)
{
    // If current value is less than
    // the current maximum
    if (maximum < (weight[node] ^ x)) {
        maximum = weight[node] ^ x;
        ans = node;
    }
    for (int to : graph[node]) {
        if (to == parent)
            continue;
        dfs(to, node);
    }
}
  
// Driver code
int main()
{
    x = 15;
  
    // 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 maximum = Integer.MIN_VALUE, x, ans;
  
    @SuppressWarnings("unchecked")
    static Vector[] graph = new Vector[100];
    static int[] weight = new int[100];
  
    // This block is executed even before main() function
    // This is necessary otherwise this program will
    // throw "NullPointerException"
    static 
    {
        for (int i = 0; i < 100; i++)
            graph[i] = new Vector<>();
    }
  
    // Function to perform dfs to find
    // the maximum xored value
    static void dfs(int node, int parent) 
    {
  
        // If current value is less than
        // the current maximum
        if (maximum < (weight[node] ^ x)) 
        {
            maximum = weight[node] ^ x;
            ans = node;
        }
        for (int to : graph[node]) 
        {
            if (to == parent)
                continue;
            dfs(to, node);
        }
    }
  
    // Driver Code
    public static void main(String[] args) 
    {
        x = 15;
  
        // 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

Python3

# Python3 implementation of the approach
import sys
maximum = -sys.maxsize - 1
graph = [[0 for i in range(100)]
            for j in range(100)]
weight = [0 for i in range(100)]
ans = []
  
# Function to perform dfs to find
# the maximum xored value
def dfs(node, parent):
    global maximum
      
    # If current value is less than
    # the current maximum
    if (maximum < (weight[node] ^ x)):
        maximum = weight[node] ^ x
        ans.append(node)
          
    for to in graph[node]:
        if (to == parent):
            continue
        dfs(to, node)
          
# Driver code
if __name__ == '__main__':
    x = 15
  
    # 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[0])
      
# This code is contributed by
# Surendra_Gangwar

C#

// C# implementation of the approach 
using System; 
using System.Collections.Generic; 
  
class GFG 
{ 
  
static int maximum = int.MinValue, x, 
ans = int.MaxValue; 
  
static List> graph = new List>(); 
static List weight = new List(); 
  
  
// Function to perform dfs to find 
// the maximum value 
static void dfs(int node, int parent) 
{ 
    // If current value is less than
    // the current maximum
    if (maximum < (weight[node] ^ x)) 
    {
        maximum = weight[node] ^ x;
        ans = node;
    } 
          
    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() 
{ 
    x = 15; 
  
    // 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.Write( 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，树的每个节点都处理一次，因此，如果树中总共有N个节点，则由于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。