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1.[226. 翻转二叉树](https://leetcode.cn/problems/symmetric-tree/)

给你一棵二叉树的根节点 root ，翻转这棵二叉树，并返回其根节点。

示例 1：

输入：root = [4,2,7,1,3,6,9]
输出：[4,7,2,9,6,3,1]

示例 2：

输入：root = [2,1,3]
输出：[2,3,1]

解题思路：递归和层序遍历都可以

# 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
from collections import deque
class Solution:def invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:# 方法一：这里递归的方法应该是最容易写的# if not root: return# root.left, root.right = self.invertTree(root.right), self.invertTree(root.left)# return root# 方法二：层序遍历应该也是可以的，常见思路if not root:return queue = deque()queue.append(root)while queue:for i in range(len(queue)):node = queue.popleft()node.left, node.right = node.right, node.leftif node.left: queue.append(node.left)if node.right: queue.append(node.right)return root

2.101. 对称二叉树

给你一个二叉树的根节点 root ， 检查它是否轴对称。

示例 1：

输入：root = [1,2,2,3,4,4,3]
输出：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:def istreessymetric(node1, node2):if not node1 and node2:return Falseif not node2 and node1:return Falseif not node1 and not node2:return Truereturn node1.val == node2.val and istreessymetric(node1.left, node2.right) and \istreessymetric(node1.right, node2.left)if not root:return Truereturn istreessymetric(root.left, root.right)

Better Implementation

# 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:def is_mirror(n1, n2):# 如果两个节点都为空，则对称if not n1 and not n2:return True# 一个为空另一个非空，或值不相等则不对称if not n1 or not n2 or n1.val != n2.val:return False# 递归检查子树：左对右，右对左return is_mirror(n1.left, n2.right) and is_mirror(n1.right, n2.left)# 处理空树情况，并检查左右子树是否互为镜像return is_mirror(root.left, root.right) if root else True