计算给定树的加权字符串为回文的节点

给定一棵树，以及所有节点的权重(以字符串的形式)，任务是计算权重为回文的节点。

例子：

输入： 输出： 3仅节点2、3和5的权重是回文。

方法：在树上执行dfs，对于每个节点，检查其字符串是否为回文。如果是，则增加计数。

执行：

C++

// C++ implementation of the approach
#include 
using namespace std;
  
int cnt = 0;
  
vector graph[100];
vector weight(100);
  
// Function that returns true
// if x is a palindrome
bool isPalindrome(string x)
{
    int n = x.size();
    for (int i = 0; i < n / 2; i++) {
        if (x[i] != x[n - 1 - i])
            return false;
    }
    return true;
}
  
// Function to perform dfs
void dfs(int node, int parent)
{
  
    // Weight of the current node
    string x = weight[node];
  
    // If the weight is a palindrome
    if (isPalindrome(x))
        cnt += 1;
  
    for (int to : graph[node]) {
        if (to == parent)
            continue;
        dfs(to, node);
    }
}
  
// Driver code
int main()
{
  
    // Weights of the node
    weight[1] = "abc";
    weight[2] = "aba";
    weight[3] = "bcb";
    weight[4] = "moh";
    weight[5] = "aa";
  
    // 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 << cnt;
  
    return 0;
}

Java

// Java implementation of the approach 
import java.util.*;
  
class GFG
{
  
static int cnt = 0; 
  
static Vector> graph = new Vector>(); 
static Vector weight = new Vector(); 
  
// Function that returns true 
// if x is a palindrome 
static boolean isPalindrome(String x) 
{ 
    int n = x.length(); 
    for (int i = 0; i < n / 2; i++)
    { 
        if (x.charAt(i) != x.charAt(n - 1 - i)) 
            return false; 
    } 
    return true; 
} 
  
// Function to perform dfs 
static void dfs(int node, int parent) 
{ 
  
    // Weight of the current node 
    String x = weight.get(node); 
      
  
    // If the weight is a palindrome 
    if (isPalindrome(x)) 
        cnt += 1; 
  
    for (int i=0;i());
      
    // 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( cnt); 
}
}
  
// This code is contributed by Arnab Kundu

Python3

# Python3 implementation of the approach
cnt = 0
  
graph = [0] * 100
for i in range(100):
    graph[i] = []
  
weight = ["0"] * 100
  
# Function that returns true
# if x is a palindrome
def isPalindrome(x):
    n = len(x)
  
    for i in range(0, n // 2):
        if x[i] != x[n - 1 - i]:
            return False
  
    return True
  
# Function to perform dfs
def dfs(node, parent):
    global cnt
  
    # Weight of the current node
    x = weight[node]
  
    # If the weight is a palindrome
    if (isPalindrome(x)):
        cnt += 1
  
    for to in graph[node]:
        if to == parent:
            continue
        dfs(to, node)
  
# Driver Code
if __name__ == "__main__":
  
    # Weights of the node
    weight[0] = ""
    weight[1] = "abc"
    weight[2] = "aba"
    weight[3] = "bcb"
    weight[4] = "moh"
    weight[5] = "aa"
  
    # Edges of the tree
    graph[1].append(2)
    graph[2].append(3)
    graph[2].append(4)
    graph[1].append(5)
  
    dfs(1, 1)
  
    print(cnt)
  
# This code is contributed by
# sanjeev2552

C#

// C# implementation of the approach 
using System;
using System.Collections.Generic; 
  
class GFG
{
  
    static int cnt = 0; 
      
    static List> graph = new List>(); 
    static List weight = new List(); 
      
    // Function that returns true
    // if x is a palindrome
    static bool isPalindrome(string x)
    {
        int n = x.Length;
        for (int i = 0; i < n / 2; i++)
        {
            if (x[i] != x[n - 1 - i])
                return false;
        }
        return true;
    }
      
    // Function to perform dfs 
    static void dfs(int node, int parent) 
    { 
      
        // Weight of the current node 
        String x = weight[node]; 
      
        // If the weight is a palindrome 
        if (isPalindrome(x)) 
            cnt += 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( ""); 
        weight.Add( "abc"); 
        weight.Add( "aba"); 
        weight.Add( "bcb"); 
        weight.Add( "moh"); 
        weight.Add( "aa"); 
          
        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( cnt); 
      
    } 
}
  
// This code has been contributed by 29AjayKumar

输出：

3

复杂度分析：

• 时间复杂度： O(N * Len)其中，Len是给定树中节点的加权字符串的最大长度。
  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)。同样，每个节点的处理都涉及遍历该节点的加权字符串一次，从而增加了O(Len)的复杂度，其中Len是加权字符串的长度。因此，总时间复杂度为O(N * Len)。
• 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。