计算给定树的节点，其加权字符串是回文

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：

Input: 

Output: 3
Only the weights of the nodes 2, 3 and 5 are palindromes.

方法：在树上执行 dfs，对于每个节点，检查它的字符串是否为回文。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

Javascript

  
// Javascript implementation of the approach
let cnt = 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 x is a palindrome
function isPalindrome(x)
{
    let n = x.length;
    for(let i = 0; i < n / 2; i++)
    {
        if (x[i] != x[n - 1 - i])
            return false;
    }
    return true;
}
 
// Function to perform dfs
function dfs(node, parent)
{
 
    // Weight of the current node
    let x = weight[node];
 
    // If the weight is a palindrome
    if (isPalindrome(x))
        cnt += 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] = "abc";
weight[2] = "aba";
weight[3] = "bcb";
weight[4] = "moh";
weight[5] = "aa";
 
// 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(cnt);
 
// This code is contributed by Dharanendra L V.
      

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 的复杂性是 O(N)。此外，每个节点的处理都涉及遍历该节点的加权字符串一次，因此增加了 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。

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