0101-symmetric-tree.py

# time complexity: O(n)
# space complexity: O(n)
from typing import Optional


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:
        def dfs(node1: Optional[TreeNode], node2: Optional[TreeNode]):
            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 dfs(node1.left, node2.right) and dfs(node1.right, node2.left)
        return dfs(root, root)


root = TreeNode(1)
root.left = TreeNode(2)
root.right = TreeNode(2)
root.left.left = TreeNode(3)
root.left.right = TreeNode(4)
root.right.left = TreeNode(4)
root.right.right = TreeNode(3)
print(Solution().isSymmetric(root))