101. Symmetric Tree | LeetCode Using Python

In this video, I solve the LeetCode problem “101. Symmetric Tree” using both recursive and iterative DFS approaches in Python. I walk through each method to check if a binary tree is symmetric around its center, demonstrating the algorithm’s effectiveness in ensuring mirrored node values and structure. Below is the code used to solve this problem:

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def isSymmetric(self, root: Optional[TreeNode]) -> bool:

        #recursive approach
        #Time Complexity: O(n) - all nodes are visite donce
        #Space Complexity: O(H) H being height - in a balance tree --> it is a O(logn) ; in a skewed tree --> it is O(n)

        #let's make a helper function
        def symmetric_check(node1, node2):
            #Base case
            if node1 is None and node2 is None:
                return True

            if node1 is None or node2 is None:
                return False

            return (node1.val == node2.val) and symmetric_check(node1.left, node2.right) and symmetric_check(node1.right, node2.left)

        return symmetric_check(root, root)


'''
        #iteratively approach - Has more advantage for deeper tree than recursive
        #Time Complexity: O(n) - all nodes are visite donce
        #Space Complexity: O(H) H being height - in a balance tree --> it is a O(logn) ; in a skewed tree --> it is O(n)

        if root is None:
            return None

        stack = [(root.left, root.right)] #Tuple in our stack

        while stack:
            node1, node2 = stack.pop()

            if node1 is None and node2 is None:
                continue

            if node1 is None or node2 is None:
                return False

            if node1.val != node2.val:
                return False

            stack.append((node1.left,node2.right))
            stack.append((node1.right, node2.left))
        

        return True

'''