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Count of leaf nodes of the tree whose weighted string is a palindrome

  • Last Updated : 22 Jun, 2021

Given an N-ary tree, and the weights which are in the form of strings of all the nodes, the task is to count the number of leaf nodes whose weights are palindrome.

Examples:  

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.

Approach: To solve the problem mentioned above follow the steps given below:  

  • Depth First Search can be used to traverse the complete tree.
  • We will keep track of parent while traversing to avoid the visited node array.
  • Initially for every node we can set a flag and if the node have at least one child (i.e. non-leaf node) then we will reset the flag.
  • The nodes with no children are the leaf nodes. For every leaf node, we will check if it’s string is palindrome or not. If yes then increment the count.

Below is the implementation of the above approach:  

C++




// C++ implementation of the approach
 
#include <bits/stdc++.h>
using namespace std;
 
int cnt = 0;
 
vector<int> graph[100];
vector<string> 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<Integer> []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<Integer>();
         
    // 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<int> []graph = new List<int>[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<int>();
         
    // 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

Javascript




<script>
 
    // JavaScript implementation of the approach
     
    let cnt = 0;
  
    let graph = new Array(100);
    let weight = new Array(100);
 
    // Function that returns true
    // if x is a palindrome
    function isPalindrome(x)
    {
        let n = x.length;
        for (let i = 0; i < parseInt(n / 2, 10); i++)
        {
            if (x[i] != x[n - 1 - i])
                return false;
        }
        return true;
    }
 
    // Function to perform DFS on the tree
    function dfs(node, parent)
    {
        let flag = 1;
 
        // Iterating the children of current node
        for (let to = 0; to < graph[node].length; to++)
        {
 
            // There is at least a child
            // of the current node
            if (graph[node][to] == parent)
                continue;
            flag = 0;
            dfs(graph[node][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
            let x = weight[node];
 
            // If the weight is a palindrome
            if (isPalindrome(x))
                cnt += 1;
        }
    }
     
    for(let i = 0; i < graph.length;i++)
        graph[i] = [];
          
    // 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(2);
    graph[2].push(3);
    graph[2].push(4);
    graph[1].push(5);
  
    dfs(1, 1);
  
    document.write(cnt);
     
</script>
Output: 
2

 




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