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# Lexicographically smallest Palindromic Path in a Binary Tree

Given a Binary Tree with each node representing an alphabet, the task is to find lexicographically smallest palindromic root to leaf path. If no palindromic path exists, print “No Palindromic Path exists”.

Examples:

Input:
a
/   \
c      b
/ \    /  \
a   g b   x
\
a
Output:
abba
Explanation:
There were total 4 root to leaf paths out of which 2 paths(i.e., “aca” and “abba”) were palindromic paths but as “abba” is lexicographically smaller, print “abba” as output.

Input:
a
/   \
z     k
/ \   / \
s   e k   u
\
e
Output:
No Palindromic Path exists

Approach: Follow the steps to solve the problem

1. The main idea is to use Preorder Traversal
2. Traverse the tree in Preorder fashion.
3. Keep storing the node values in a string.
4. As soon as a leaf node is reached, check if the string formed from a root to leaf path is a palindrome or not.
5. If it’s a palindrome store it in a variable only if it’s lexicographically the smallest palindromic path.
6. Print the palindrome, if it exists.

Below is the implementation of the above approach:

## C++

 `// C++ program``// for the above approach` `#include ``using` `namespace` `std;` `// Struct binary tree node``struct` `Node {``    ``char` `data;``    ``Node *left, *right;``};` `// Function to create a new node``Node* newNode(``char` `data)``{``    ``Node* temp = ``new` `Node();``    ``temp->data = data;``    ``temp->left = temp->right = NULL;``    ``return` `temp;``}` `// Function to check if the``// string is palindrome or not``bool` `checkPalindrome(string s)``{``    ``int` `low = 0, high = (``int``)s.size() - 1;` `    ``while` `(low < high) {` `        ``if` `(s[low] != s[high])``            ``return` `false``;` `        ``low++;``        ``high--;``    ``}` `    ``return` `true``;``}` `// Function to find the lexicographically``// smallest palindromic path in the Binary Tree``void` `lexicographicallySmall(Node* root, string s,``                            ``string& finalAns)``{``    ``// Base case``    ``if` `(root == NULL)``        ``return``;` `    ``// Append current node's``    ``// data to the string``    ``s += root->data;` `    ``// Check if a node is leaf or not``    ``if` `(!root->left and !root->right) {` `        ``if` `(checkPalindrome(s)) {` `            ``// Check for the 1st``            ``// Palindromic Path``            ``if` `(finalAns == ``"\$"``)``                ``finalAns = s;` `            ``// Store lexicographically the``            ``// smallest palindromic path``            ``else``                ``finalAns = min(finalAns, s);``        ``}` `        ``return``;``    ``}` `    ``// Recursively traverse left subtree``    ``lexicographicallySmall(root->left,``                           ``s, finalAns);` `    ``// Recursively traverse right subtree``    ``lexicographicallySmall(root->right,``                           ``s, finalAns);``}` `// Function to get smallest``// lexicographical palindromic``// path``void` `getPalindromePath(Node* root)``{``    ``// Variable which stores``    ``// the final result``    ``string finalAns = ``"\$"``;` `    ``// Function call to compute``    ``// lexicographically smallest``    ``// palindromic Path``    ``lexicographicallySmall(root, ``""``,``                           ``finalAns);` `    ``if` `(finalAns == ``"\$"``)``        ``cout << ``"No Palindromic Path exists"``;` `    ``else``        ``cout << finalAns;``}``// Driver Code``int` `main()``{``    ``// Construct binary tree``    ``Node* root = newNode(``'a'``);``    ``root->left = newNode(``'c'``);``    ``root->left->left = newNode(``'a'``);``    ``root->left->right = newNode(``'g'``);``    ``root->right = newNode(``'b'``);``    ``root->right->left = newNode(``'b'``);``    ``root->right->right = newNode(``'x'``);``    ``root->right->left->right = newNode(``'a'``);` `    ``getPalindromePath(root);` `    ``return` `0;``}`

## Java

 `// Java program``// for the above approach``import` `java.util.*;``class` `GFG``{``static` `String finalAns=``""``;``  ` `// Struct binary tree node``static` `class` `Node``{``    ``char` `data;``    ``Node left, right;``};` `// Function to create a new node``static` `Node newNode(``char` `data)``{``    ``Node temp = ``new` `Node();``    ``temp.data = data;``    ``temp.left = temp.right = ``null``;``    ``return` `temp;``}` `// Function to check if the``// String is palindrome or not``static` `boolean` `checkPalindrome(String s)``{``    ``int` `low = ``0``, high = (``int``)s.length() - ``1``;``    ``while` `(low < high)``    ``{``        ``if` `(s.charAt(low) != s.charAt(high))``            ``return` `false``;``        ``low++;``        ``high--;``    ``}``    ``return` `true``;``}` `// Function to find the lexicographically``// smallest palindromic path in the Binary Tree``static` `void` `lexicographicallySmall(Node root, String s)``{``  ` `    ``// Base case``    ``if` `(root == ``null``)``        ``return``;` `    ``// Append current node's``    ``// data to the String``    ``s += root.data;` `    ``// Check if a node is leaf or not``    ``if` `(root.left == ``null` `&& root.right == ``null``)``    ``{``        ``if` `(checkPalindrome(s))``        ``{` `            ``// Check for the 1st``            ``// Palindromic Path``            ``if` `(finalAns == ``"\$"``)``                ``finalAns = s;` `            ``// Store lexicographically the``            ``// smallest palindromic path``            ``else``                ``finalAns = finalAns.compareTo(s) <= ``0` `? finalAns:s;``        ``}``        ``return``;``    ``}` `    ``// Recursively traverse left subtree``    ``lexicographicallySmall(root.left,``                           ``s);` `    ``// Recursively traverse right subtree``    ``lexicographicallySmall(root.right,``                           ``s);``}` `// Function to get smallest``// lexicographical palindromic``// path``static` `void` `getPalindromePath(Node root)``{``  ` `    ``// Variable which stores``    ``// the final result``    ``finalAns = ``"\$"``;` `    ``// Function call to compute``    ``// lexicographically smallest``    ``// palindromic Path``    ``lexicographicallySmall(root, ``""``);``    ``if` `(finalAns == ``"\$"``)``        ``System.out.print(``"No Palindromic Path exists"``);``    ``else``        ``System.out.print(finalAns);``}``  ` `// Driver Code``public` `static` `void` `main(String[] args)``{``  ` `    ``// Construct binary tree``    ``Node root = newNode(``'a'``);``    ``root.left = newNode(``'c'``);``    ``root.left.left = newNode(``'a'``);``    ``root.left.right = newNode(``'g'``);``    ``root.right = newNode(``'b'``);``    ``root.right.left = newNode(``'b'``);``    ``root.right.right = newNode(``'x'``);``    ``root.right.left.right = newNode(``'a'``);``    ``getPalindromePath(root);``}``}` `// This code is contributed by 29AjayKumar`

## Python3

 `# Python3 program``# for the above approach` `# Struct binary tree node``class` `Node:``    ``def` `__init__(``self``, d):``        ``self``.data ``=` `d``        ``self``.left ``=` `None``        ``self``.right ``=` `None` `# Function to check if the``# is palindrome or not``def` `checkPalindrome(s):``    ``low, high ``=` `0``, ``len``(s) ``-` `1``    ``while` `(low < high):``        ``if` `(s[low] !``=` `s[high]):``            ``return` `False``        ``low ``+``=` `1``        ``high ``-``=` `1``    ``return` `True` `# Function to find the lexicographically``# smallest palindromic path in the Binary Tree``def` `lexicographicallySmall(root, s):``    ``global` `finalAns``    ` `    ``# Base case``    ``if` `(root ``=``=` `None``):``        ``return` `    ``# Append current node's``    ``# data to the string``    ``s ``+``=` `root.data` `    ``# Check if a node is leaf or not``    ``if` `(``not` `root.left ``and` `not` `root.right):``        ``if` `(checkPalindrome(s)):` `            ``# Check for the 1st``            ``# Palindromic Path``            ``if` `(finalAns ``=``=` `"\$"``):``                ``finalAns ``=` `s` `            ``# Store lexicographically the``            ``# smallest palindromic path``            ``else``:``                ``finalAns ``=` `min``(finalAns, s)``        ``return` `    ``# Recursively traverse left subtree``    ``lexicographicallySmall(root.left, s)` `    ``# Recursively traverse right subtree``    ``lexicographicallySmall(root.right, s)` `# Function to get smallest``# lexicographical palindromic``# path``def` `getPalindromePath(root):``    ``global` `finalAns``    ` `    ``# Variable which stores``    ``# the final result``    ``finalAns ``=` `"\$"` `    ``# Function call to compute``    ``# lexicographically smallest``    ``# palindromic Path``    ``lexicographicallySmall(root, "")``    ``if` `(finalAns ``=``=` `"\$"``):``        ``print``(``"No Palindromic Path exists"``)``    ``else``:``        ``print``(finalAns)` `        ``# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ``finalAns ``=` `""``    ` `    ``# Construct binary tree``    ``root ``=` `Node(``'a'``)``    ``root.left ``=` `Node(``'c'``)``    ``root.left.left ``=` `Node(``'a'``)``    ``root.left.right ``=` `Node(``'g'``)``    ``root.right ``=` `Node(``'b'``)``    ``root.right.left ``=` `Node(``'b'``)``    ``root.right.right ``=` `Node(``'x'``)``    ``root.right.left.right ``=` `Node(``'a'``)` `    ``getPalindromePath(root)` `    ``# This code is contributed by mohit kumar 29.`

## C#

 `// C# program``// for the above approach``using` `System;` `public` `class` `GFG``{``static` `String finalAns = ``""``;``  ` `// Struct binary tree node``class` `Node``{``    ``public` `char` `data;``    ``public` `Node left, right;``};` `// Function to create a new node``static` `Node newNode(``char` `data)``{``    ``Node temp = ``new` `Node();``    ``temp.data = data;``    ``temp.left = temp.right = ``null``;``    ``return` `temp;``}` `// Function to check if the``// String is palindrome or not``static` `bool` `checkPalindrome(String s)``{``    ``int` `low = 0, high = (``int``)s.Length - 1;``    ``while` `(low < high)``    ``{``        ``if` `(s[low] != s[high])``            ``return` `false``;``        ``low++;``        ``high--;``    ``}``    ``return` `true``;``}` `// Function to find the lexicographically``// smallest palindromic path in the Binary Tree``static` `void` `lexicographicallySmall(Node root, String s)``{``  ` `    ``// Base case``    ``if` `(root == ``null``)``        ``return``;` `    ``// Append current node's``    ``// data to the String``    ``s += root.data;` `    ``// Check if a node is leaf or not``    ``if` `(root.left == ``null` `&& root.right == ``null``)``    ``{``        ``if` `(checkPalindrome(s))``        ``{` `            ``// Check for the 1st``            ``// Palindromic Path``            ``if` `(finalAns == ``"\$"``)``                ``finalAns = s;` `            ``// Store lexicographically the``            ``// smallest palindromic path``            ``else``                ``finalAns = finalAns.CompareTo(s) <= 0 ? finalAns:s;``        ``}``        ``return``;``    ``}` `    ``// Recursively traverse left subtree``    ``lexicographicallySmall(root.left,``                           ``s);` `    ``// Recursively traverse right subtree``    ``lexicographicallySmall(root.right,``                           ``s);``}` `// Function to get smallest``// lexicographical palindromic``// path``static` `void` `getPalindromePath(Node root)``{``  ` `    ``// Variable which stores``    ``// the readonly result``    ``finalAns = ``"\$"``;` `    ``// Function call to compute``    ``// lexicographically smallest``    ``// palindromic Path``    ``lexicographicallySmall(root, ``""``);``    ``if` `(finalAns == ``"\$"``)``        ``Console.Write(``"No Palindromic Path exists"``);``    ``else``        ``Console.Write(finalAns);``}``  ` `// Driver Code``public` `static` `void` `Main(String[] args)``{``  ` `    ``// Construct binary tree``    ``Node root = newNode(``'a'``);``    ``root.left = newNode(``'c'``);``    ``root.left.left = newNode(``'a'``);``    ``root.left.right = newNode(``'g'``);``    ``root.right = newNode(``'b'``);``    ``root.right.left = newNode(``'b'``);``    ``root.right.right = newNode(``'x'``);``    ``root.right.left.right = newNode(``'a'``);``    ``getPalindromePath(root);``}``}` `// This code is contributed by 29AjayKumar`

## Javascript

 ``

Output:

`abba`

Time Complexity: O(N2)
Auxiliary Space: O(N2)