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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 ``using` `namespace` `std;` `int` `cnt = 0;` `vector<``int``> graph;``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 = ``"ab"``;``    ``weight = ``"abca"``;``    ``weight = ``"axxa"``;``    ``weight = ``"geeks"``;``    ``weight = ``"aba"``;` `    ``// Edges of the tree``    ``graph.push_back(2);``    ``graph.push_back(3);``    ``graph.push_back(4);``    ``graph.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<``int``> []graph = ``new` `List<``int``>;``static` `String []weight = ``new` `String;` `// 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 = ``"ab"``;``    ``weight = ``"abca"``;``    ``weight = ``"axxa"``;``    ``weight = ``"geeks"``;``    ``weight = ``"aba"``;` `    ``// Edges of the tree``    ``graph.Add(2);``    ``graph.Add(3);``    ``graph.Add(4);``    ``graph.Add(5);` `    ``dfs(1, 1);` `    ``Console.Write(cnt);``}``}` `// This code is contributed by amal kumar choubey`

## Javascript

 ``

Output:

`2`

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