Problem

Given the root of a binary tree, check whether it is a mirror of itself (i.e., symmetric around its center).

Example 1:

!https://assets.leetcode.com/uploads/2021/02/19/symtree1.jpg

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Input: root = [1,2,2,3,4,4,3]
Output: true

Example 2:

!https://assets.leetcode.com/uploads/2021/02/19/symtree2.jpg

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Input: root = [1,2,2,null,3,null,3]
Output: false

Constraints:

  • The number of nodes in the tree is in the range [1, 1000].
  • 100 <= Node.val <= 100

Follow up:

Could you solve it both recursively and iteratively?

Solve

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# 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:
        if not root:
            return True
        
        def dfs(left,right):
            if not left and not right:
                return True
            
            if not left or not right:
                return False
            
            return left.val == right.val and dfs(left.left,right.right) and dfs(left.right, right.left)
        
        return dfs(root.left , root.right)