重新求职刷题力扣DAY15

1.[226. 翻转二叉树]( leetcode /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.left if 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 False if not node2 and node1: return False if not node1 and not node2: return True return node1.val == node2.val and istreessymetric(node1.left, node2.right) and \ istreessymetric(node1.right, node2.left) if not root: return True return 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