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Lexicographically smallest Palindromic Path in a Binary Tree

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  • Difficulty Level : Easy
  • Last Updated : 07 Mar, 2022

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 <bits/stdc++.h>
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




<script>
    // Javascript program for the above approach
    let finalAns="";
     
    // Struct binary tree node
    class Node
    {
        constructor(data) {
           this.left = null;
           this.right = null;
           this.data = data;
        }
    }
 
    // Function to create a new node
    function newNode(data)
    {
        let temp = new Node(data);
        return temp;
    }
 
    // Function to check if the
    // String is palindrome or not
    function checkPalindrome(s)
    {
        let low = 0, high = 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
    function lexicographicallySmall(root, 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.localeCompare(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
    function getPalindromePath(root)
    {
 
        // Variable which stores
        // the final result
        finalAns = "$";
 
        // Function call to compute
        // lexicographically smallest
        // palindromic Path
        lexicographicallySmall(root, "");
        if (finalAns == "$")
            document.write("No Palindromic Path exists");
        else
            document.write(finalAns);
    }
     
    // Construct binary tree
    let 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 suresh07.
</script>


 
 

Output: 

abba

 

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

 


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