101. Symmetric Tree

Easy LeetCode

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

Example 1

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

Output: true

Example 2

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

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How to solve the problem

Recursion (DFS Only Postorder Traverse)

# 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 root is None:
            return True
        def isMirror(left, right):
            if not left and not right:
                return True
            if not left or not right:
                return False
            if left.val != right.val:
                return False
            return isMirror(left.left, right.right) and isMirror(left.right, right.left)
        return isMirror(root.left, root.right)

Complexity

• Time complexity: O(n), n == number of nodes
• Space complexity: O(n)