https://leetcode-cn.com/submissions/detail/305476984/ (opens new window) 需要注意的点 1.迭代 2.递归

# 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
        
        left = root.left
        right = root.right

        if (not left and right) or (not right and left):
            return False

        q = [left, right]

        while q:
            n1 = q.pop(0)
            n2 = q.pop(0)

            if not n1 and not n2:
                continue

            if not n1 or not n2:
                return False
            
            if n1.val != n2.val:
                return False
            
            q.append(n1.left)
            q.append(n2.right)
            q.append(n1.right)
            q.append(n2.left)

        return True

# 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 reverseTree(left: TreeNode, right: TreeNode)-> bool:
            if not left or not right:
                return left == right
            
            if left.val != right.val:
                return False

            return (reverseTree(left.left, right.right) and reverseTree(left.right, right.left)) 

        return reverseTree(root.left, root.right)