Check if Inorder traversal of a Binary Tree is palindrome or not

Given a binary tree and the task if to check if its Inorder Sequence is a palindrome or not. 
Examples:

Input:

Output: True 
Explanation:
The Inorder sequence of the tree is “bbaaabb” which is a palindromic string.
Input:

Output: False 
Explanation:
The Inorder sequence of the tree is “bbdaabb” which is not a palindromic string. 

Approach:
To solve the problem, refer to the following steps: 

• Convert the given Binary Tree to a Doubly Linked List.
• This, reduces the problem to Check if a doubly linked list of characters is palindrome or not.

Below is the implementation of above approach: 

C++

// C++ Program to check if
// the Inorder sequence of a
// Binary Tree is a palindrome

#include <bits/stdc++.h>
using namespace std;

// Define node of tree
struct node {
    char val;
    node* left;
    node* right;
};

// Function to create the node
node* newnode(char i)
{
    node* temp = NULL;
    temp = new node();
    temp->val = i;
    temp->left = NULL;
    temp->right = NULL;
    return temp;
}

// Function to convert the tree
// to the double linked list
node* conv_tree(node* root,
                node* shoot)
{
    if (root->left != NULL)
        shoot = conv_tree(root->left,
                          shoot);

    root->left = shoot;

    if (shoot != NULL)
        shoot->right = root;

    shoot = root;

    if (root->right != NULL)
        shoot = conv_tree(root->right,
                          shoot);

    return shoot;
}

// Function for checking linked
// list is palindrome or not
int checkPalin(node* root)
{
    node* voot = root;

    // Count the nodes
    // form the back
    int j = 0;

    // Set the pointer at the
    // very first node of the
    // linklist and count the
    // length of the linklist
    while (voot->left != NULL) {
        j = j + 1;
        voot = voot->left;
    }
    int i = 0;

    while (i < j) {
        // Check if the value matches
        if (voot->val != root->val)
            return 0;

        else {
            i = i + 1;
            j = j - 1;
            voot = voot->right;
            root = root->left;
        }
    }

    return 1;
}

// Driver Program
int main()
{
    node* root = newnode('b');
    root->left = newnode('b');
    root->right = newnode('a');
    root->left->right = newnode('b');
    root->left->left = newnode('a');

    node* shoot = conv_tree(root, NULL);

    if (checkPalin(shoot))
        cout << "True" << endl;
    else
        cout << "False" << endl;
}

Java

// Java Program to check if
// the Inorder sequence of a
// Binary Tree is a palindrome
import java.util.*;
class GFG{

// Define node of tree
static class node
{
    char val;
    node left;
    node right;
};

// Function to create the node
static node newnode(char i)
{
    node temp = null;
    temp = new node();
    temp.val = i;
    temp.left = null;
    temp.right = null;
    return temp;
}

// Function to convert the tree
// to the double linked list
static node conv_tree(node root,
                      node shoot)
{
    if (root.left != null)
        shoot = conv_tree(root.left,
                          shoot);

    root.left = shoot;

    if (shoot != null)
        shoot.right = root;

    shoot = root;

    if (root.right != null)
        shoot = conv_tree(root.right,
                          shoot);

    return shoot;
}

// Function for checking linked
// list is palindrome or not
static int checkPalin(node root)
{
    node voot = root;

    // Count the nodes
    // form the back
    int j = 0;

    // Set the pointer at the
    // very first node of the
    // linklist and count the
    // length of the linklist
    while (voot.left != null)
    {
        j = j + 1;
        voot = voot.left;
    }
    int i = 0;

    while (i < j)
    {
        // Check if the value matches
        if (voot.val != root.val)
            return 0;

        else
        {
            i = i + 1;
            j = j - 1;
            voot = voot.right;
            root = root.left;
        }
    }
    return 1;
}

// Driver Code
public static void main(String[] args)
{
    node root = newnode('b');
    root.left = newnode('b');
    root.right = newnode('a');
    root.left.right = newnode('b');
    root.left.left = newnode('a');

    node shoot = conv_tree(root, null);

    if (checkPalin(shoot) == 1)
        System.out.print("True" + "\n");
    else
        System.out.print("False" + "\n");
}
}

// This code is contributed by 29AjayKumar

Python3

# Python3 program to check if
# the Inorder sequence of a
# Binary Tree is a palindrome

# Define node of tree
class Node:
    def __init__(self,val):

        self.val = val
        self.left = None
        self.right = None

# Function to create the node
def newnode(i):

    #temp = None;
    temp = Node(i);
    temp.val = i;
    temp.left = None;
    temp.right = None;
    return temp;

# Function to convert the tree
# to the double linked list
def conv_tree(root, shoot):

    if (root.left != None):
        shoot = conv_tree(root.left, shoot);

    root.left = shoot;

    if (shoot != None):
        shoot.right = root;

    shoot = root;

    if (root.right != None):
        shoot = conv_tree(root.right, shoot);

    return shoot;

# Function for checking linked
# list is palindrome or not
def checkPalin(root):

    voot = root;

    # Count the nodes
    # form the back
    j = 0;

    # Set the pointer at the
    # very first node of the
    # linklist and count the
    # length of the linklist
    while (voot.left != None):
        j = j + 1;
        voot = voot.left;

    i = 0;

    while (i < j):

        # Check if the value matches
        if (voot.val != root.val):
            return 0;
        else:
            i = i + 1;
            j = j - 1;
            voot = voot.right;
            root = root.left;

    return 1;

# Driver code
if __name__=='__main__':

    root = newnode('b');
    root.left = newnode('b');
    root.right = newnode('a');
    root.left.right = newnode('b');
    root.left.left = newnode('a');

    shoot = conv_tree(root, None);

    if (checkPalin(shoot)):
        print("True");
    else:
        print("False");

# This code is contributed by AbhiThakur

C#

// C# program to check if the inorder
// sequence of a Binary Tree is a
// palindrome
using System;

class GFG{

// Define node of tree
class node
{
    public char val;
    public node left;
    public node right;
};

// Function to create the node
static node newnode(char i)
{
    node temp = null;
    temp = new node();
    temp.val = i;
    temp.left = null;
    temp.right = null;

    return temp;
}

// Function to convert the tree
// to the double linked list
static node conv_tree(node root,
                      node shoot)
{
    if (root.left != null)
        shoot = conv_tree(root.left,
                          shoot);

    root.left = shoot;

    if (shoot != null)
        shoot.right = root;

    shoot = root;

    if (root.right != null)
        shoot = conv_tree(root.right,
                          shoot);

    return shoot;
}

// Function for checking linked
// list is palindrome or not
static int checkPalin(node root)
{
    node voot = root;

    // Count the nodes
    // form the back
    int j = 0;

    // Set the pointer at the
    // very first node of the
    // linklist and count the
    // length of the linklist
    while (voot.left != null)
    {
        j = j + 1;
        voot = voot.left;
    }
    int i = 0;

    while (i < j)
    {

        // Check if the value matches
        if (voot.val != root.val)
            return 0;
        else
        {
            i = i + 1;
            j = j - 1;
            voot = voot.right;
            root = root.left;
        }
    }
    return 1;
}

// Driver Code
public static void Main(String[] args)
{
    node root = newnode('b');
    root.left = newnode('b');
    root.right = newnode('a');
    root.left.right = newnode('b');
    root.left.left = newnode('a');

    node shoot = conv_tree(root, null);

    if (checkPalin(shoot) == 1)
        Console.Write("True" + "\n");
    else
        Console.Write("False" + "\n");
}
}

// This code is contributed by 29AjayKumar

Javascript

<script>
      // JavaScript program to check if the inorder
      // sequence of a Binary Tree is a
      // palindrome
      // Define node of tree
      class node {
        constructor() {
          this.val = 0;
          this.left = null;
          this.right = null;
        }
      }

      // Function to create the node
      function newnode(i) {
        var temp = null;
        temp = new node();
        temp.val = i;
        temp.left = null;
        temp.right = null;

        return temp;
      }

      // Function to convert the tree
      // to the double linked list
      function conv_tree(root, shoot) {
        if (root.left != null) shoot = conv_tree(root.left, shoot);

        root.left = shoot;

        if (shoot != null) shoot.right = root;

        shoot = root;

        if (root.right != null) shoot = conv_tree(root.right, shoot);

        return shoot;
      }

      // Function for checking linked
      // list is palindrome or not
      function checkPalin(root) {
        var voot = root;

        // Count the nodes
        // form the back
        var j = 0;

        // Set the pointer at the
        // very first node of the
        // linklist and count the
        // length of the linklist
        while (voot.left != null) {
          j = j + 1;
          voot = voot.left;
        }
        var i = 0;

        while (i < j) {
          // Check if the value matches
          if (voot.val != root.val) return 0;
          else {
            i = i + 1;
            j = j - 1;
            voot = voot.right;
            root = root.left;
          }
        }
        return 1;
      }

      // Driver Code
      var root = newnode("b");
      root.left = newnode("b");
      root.right = newnode("a");
      root.left.right = newnode("b");
      root.left.left = newnode("a");

      var shoot = conv_tree(root, null);

      if (checkPalin(shoot) == 1) document.write("True" + "<br>");
      else document.write("False" + "<br>");

      // This code is contributed by rdtank.
    </script>

Output:

True

Time Complexity: O(N) 
Auxiliary Space: O(1)