Check If Inorder Sequence of a Tree Is Palindrome in Python

Suppose we have a binary tree where each node contains a digit from 0-9, we have to check whether its in-order traversal is palindrome or not.

So, if the input is like

then the output will be True, as its inorder traversal is [2,6,10,6,2].

To solve this, we will follow these steps −

• if root is null, then
  • return True
• stack := a new stack
• curr := root
• inorder := a new list
• while stack is not empty or curr is not null, do
  • while curr is not null, do
    • push curr into stack
    • curr := left of curr
  • node := popped element from stack
  • insert value of node at the end of inorder
  • curr := right of node
• return true when inorder is same as inorder in reverse order, otherwise false.

Let us see the following implementation to get better understanding −

Example

 Live Demo

class TreeNode:
   def __init__(self, data, left = None, right = None):
      self.val = data
      self.left = left
      self.right = right
class Solution:
   def solve(self, root):
      if not root:
         return True
      stack = []
      curr = root
      inorder = []
      while stack or curr:
         while curr:
            stack.append(curr)
            curr = curr.left
         node = stack.pop() inorder.append(node.val)
         curr = node.right
      return inorder == inorder[::-1]
ob = Solution()
root = TreeNode(6)
root.left = TreeNode(2)
root.right = TreeNode(6)
root.right.left = TreeNode(10)
root.right.right = TreeNode(2)
print(ob.solve(root))

Input

root = TreeNode(6)
root.left = TreeNode(2)
root.right = TreeNode(6)
root.right.left = TreeNode(10)
root.right.right = TreeNode(2)

Output

True