📜  加权字符串为回文的树的叶节点数

📅  最后修改于: 2021-09-07 02:07:29             🧑  作者: Mango

给定一个N 叉树,以及所有节点的字符串形式的权重,任务是计算权重为回文的叶节点的数量。

例子:

Input: 
               1(ab)
              /  \
       (abca)2     5 (aba)
          /   \ 
   (axxa)3     4 (geeks)
Output: 2
Explanation: 
Only the weights of the leaf nodes
"axxa" and "aba" are palindromes.

Input: 
               1(abx)
              /  
            2(abaa) 
           /    
          3(amma)  
Output: 1
Explanation: 
Only the weight of the leaf
node "amma" is palindrome.

方法:解决上述问题,请按照以下步骤操作:

  • 深度优先搜索可用于遍历整个树。
  • 我们将在遍历时跟踪父节点以避免访问的节点数组。
  • 最初,我们可以为每个节点设置一个标志,如果该节点至少有一个子节点(即非叶节点),那么我们将重置该标志。
  • 没有孩子的节点是叶节点。对于每个叶节点,我们将检查它的字符串是否为回文。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 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 on the tree
void dfs(int node, int parent)
{
    int flag = 1;
 
    // Iterating the children of current node
    for (int to : graph[node]) {
 
        // There is at least a child
        // of the current node
        if (to == parent)
            continue;
        flag = 0;
        dfs(to, node);
    }
 
    // Current node is connected to only
    // its parent i.e. it is a leaf node
    if (flag == 1) {
        // Weight of the current node
        string x = weight[node];
 
        // If the weight is a palindrome
        if (isPalindrome(x))
            cnt += 1;
    }
}
 
// Driver code
int main()
{
 
    // Weights of the node
    weight[1] = "ab";
    weight[2] = "abca";
    weight[3] = "axxa";
    weight[4] = "geeks";
    weight[5] = "aba";
 
    // 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[100];
static String []weight = new String[100];
 
// 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 on the tree
static void dfs(int node, int parent)
{
    int flag = 1;
 
    // Iterating the children of current node
    for (int to : graph[node])
    {
 
        // There is at least a child
        // of the current node
        if (to == parent)
            continue;
        flag = 0;
        dfs(to, node);
    }
 
    // Current node is connected to only
    // its parent i.e. it is a leaf node
    if (flag == 1)
    {
        // Weight of the current node
        String x = weight[node];
 
        // If the weight is a palindrome
        if (isPalindrome(x))
            cnt += 1;
    }
}
 
// Driver code
public static void main(String[] args)
{
    for(int i = 0; i < graph.length;i++)
        graph[i] = new Vector();
         
    // Weights of the node
    weight[1] = "ab";
    weight[2] = "abca";
    weight[3] = "axxa";
    weight[4] = "geeks";
    weight[5] = "aba";
 
    // 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.print(cnt);
}
}
 
// This code is contributed by amal kumar choubey


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: str) -> bool:
     
    n = len(x)
     
    for i in range(n // 2):
        if (x[i] != x[n - 1 - i]):
            return False
 
    return True
 
# Function to perform DFS on the tree
def dfs(node: int, parent: int) -> None:
     
    global cnt, graph, weight
 
    flag = 1
 
    # Iterating the children of current node
    for to in graph[node]:
 
        # There is at least a child
        # of the current node
        if (to == parent):
            continue
         
        flag = 0
        dfs(to, node)
 
    # Current node is connected to only
    # its parent i.e. it is a leaf node
    if (flag == 1):
         
        # Weight of the current node
        x = weight[node]
 
        # If the weight is a palindrome
        if (isPalindrome(x)):
            cnt += 1
 
# Driver code
if __name__ == "__main__":
 
    # Weights of the node
    weight[1] = "ab"
    weight[2] = "abca"
    weight[3] = "axxa"
    weight[4] = "geeks"
    weight[5] = "aba"
 
    # 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[100];
static String []weight = new String[100];
 
// 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 on the tree
static void dfs(int node, int parent)
{
    int flag = 1;
 
    // Iterating the children of
    // current node
    foreach (int to in graph[node])
    {
 
        // There is at least a child
        // of the current node
        if (to == parent)
            continue;
        flag = 0;
        dfs(to, node);
    }
 
    // Current node is connected to only
    // its parent i.e. it is a leaf node
    if (flag == 1)
    {
         
        // Weight of the current node
        String x = weight[node];
 
        // If the weight is a palindrome
        if (isPalindrome(x))
            cnt += 1;
    }
}
 
// Driver code
public static void Main(String[] args)
{
    for(int i = 0; i < graph.Length; i++)
       graph[i] = new List();
         
    // Weights of the node
    weight[1] = "ab";
    weight[2] = "abca";
    weight[3] = "axxa";
    weight[4] = "geeks";
    weight[5] = "aba";
 
    // 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(cnt);
}
}
 
// This code is contributed by amal kumar choubey


输出:
2


如果您想与行业专家一起参加直播课程,请参阅Geeks Classes Live