力扣(2024.08.08)

  1. 101:对称二叉树
python 复制代码
# 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 dfs(node, level, res):
            if len(res) == level:
                res.append([])
            if not node:
                res[level].append('None')
            else:
                res[level].append(node.val)
                dfs(node.left, level + 1, res)
                dfs(node.right, level + 1, res)

        res = []
        level = 0
        dfs(root, level, res)
        for i in res:
            if i[::-1] != i:
                return False
        return True
  1. 100:相同的树
python 复制代码
# 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 isSameTree(self, p: Optional[TreeNode], q: Optional[TreeNode]) -> bool:
        def dfs(node, level, res):
            if len(res) == level:
                res.append([])
            if not node:
                res[level].append('None')
            else:
                res[level].append(node.val)
                dfs(node.left, level + 1, res)
                dfs(node.right, level + 1, res)

        res1 = []
        res2 = []
        level = 0
        dfs(p, level, res1)
        dfs(q, level, res2)
        return res1 == res2
  1. 559:N叉树的最大深度
python 复制代码
"""
# Definition for a Node.
class Node:
    def __init__(self, val=None, children=None):
        self.val = val
        self.children = children
"""

class Solution:
    def maxDepth(self, root: 'Node') -> int:
        def dfs(node, level, res):
            if not node:
                return
            if len(res) == level:
                res.append([])
            res[level].append(node.val)
            for child in node.children:
                dfs(child, level + 1, res)
        
        res = []
        level = 0
        dfs(root, level, res)
        return len(res)
  1. 111:二叉树的最小深度
python 复制代码
# 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 minDepth(self, root: Optional[TreeNode]) -> int:
        def dfs(node):
            if not node:
                return 0
            if node.left and not node.right:
                return 1 + dfs(node.left)
            elif not node.left and node.right:
                return 1 + dfs(node.right)
            elif not node.left and not node.right:
                return 1
            elif node.left and node.right:
                return 1 + min(dfs(node.left), dfs(node.right))

        res = dfs(root)
        return res
相关推荐
Jerry4 小时前
LeetCode 189. 轮转数组
算法
Jerry5 小时前
LeetCode 739. 每日温度
算法
tkevinjd8 小时前
MiniCode 项目详解6:原项目控制系统的10个缺陷(已修复)
python·llm·agent
dNGUZ7UGj9 小时前
10分钟完成第一个Python小游戏
python·django
2601_954526759 小时前
【工业传感与算法实战】温漂补偿与零点抗漂破局:基于二阶多项式拟合的 C/C++ 边缘校准算法,深度拆解“压力变送器什么牌子好”的技术硬指标
c语言·c++·算法
没有梦想的咸鱼185-1037-16639 小时前
AI-Python机器学习与深度学习技术:CNN/Transformer/扩散模型、SHAP可解释及Hermes智能体自动化
人工智能·python·深度学习·机器学习·chatgpt·cnn·transformer
霸道流氓气质10 小时前
SpringBoot中通用工具类库(Utils)封装与使用实践
spring boot·后端·python
叩码以求索10 小时前
浅谈:算法萌新如何高效刷题应对面试(一)
算法·面试·职场和发展
Draina11 小时前
CBC填充预言攻击-CBC Padding Oracle Crypto Attack
python·安全·web安全·网络安全·密码学·安全性测试
c2385612 小时前
把 C++ 内存分配拆透:new 与 malloc 的三层血缘
开发语言·c++·算法