力扣(2024.08.07)

  1. 637:二叉树的层平均值
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 averageOfLevels(self, root: Optional[TreeNode]) -> List[float]:
        def dfs(node, level, res):
            if not node:
                return
            if len(res) == level:
                res.append([])
            res[level].append(node.val)
            dfs(node.left, level + 1, res)
            dfs(node.right, level + 1, res)

        res = []
        level = 0
        dfs(root, level, res)

        final_res = []
        for i in res:
            final_res.append(sum(i) / len(i))
        return final_res
  1. 429:N叉树的层序遍历
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 levelOrder(self, root: 'Node') -> List[List[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 res       
  1. 515:在每个树行中找最大值
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 largestValues(self, root: Optional[TreeNode]) -> List[int]:
        def dfs(node, level, res):
            if not node:
                return
            if len(res) == level:
                res.append([])
            res[level].append(node.val)
            dfs(node.left, level + 1, res)
            dfs(node.right, level + 1, res)

        res = []
        level = 0
        dfs(root, level, res)

        final_res = []
        for i in res:
            final_res.append(max(i))
        return final_res
  1. 104:二叉树的最大深度
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 maxDepth(self, root: Optional[TreeNode]) -> int:
        def dfs(node, level, res):
            if not node:
                return
            if len(res) == level:
                res.append([])
            res[level].append(node.val)
            dfs(node.left, level + 1, res)
            dfs(node.right, level + 1, res)

        res = []
        level = 0
        dfs(root, level, res)
        return len(res)
  1. 226:翻转二叉树
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 invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
        def dfs(node):
            if not node:
                return
            node.left, node.right = node.right, node.left
            dfs(node.left)
            dfs(node.right)

        dfs(root)
        return root
相关推荐
位东风1 分钟前
希尔排序(Shell Sort)详解
算法·排序算法
Learner13 分钟前
Python异常处理
java·前端·python
AI科技星15 分钟前
光速飞行器动力学方程的第一性原理推导、验证与范式革命
数据结构·人工智能·线性代数·算法·机器学习·概率论
橘颂TA17 分钟前
【剑斩OFFER】算法的暴力美学——leetCode 946 题:验证栈序列
c++·算法·leetcode·职场和发展·结构与算法
闻缺陷则喜何志丹19 分钟前
【状态机动态规划】3686. 稳定子序列的数量|1969
c++·算法·动态规划·力扣·状态机动态规划
hui函数27 分钟前
Python系列Bug修复|如何解决 pip install 安装报错 Backend ‘setuptools.build_meta’ 不可用 问题
python·bug·pip
谢的2元王国28 分钟前
prompt工程逐渐成为工作流的重要一部分:以下是一套多节点新闻处理外加事实增强的文章报告日志记录
python
寻星探路30 分钟前
【算法通关】双指针技巧深度解析:从基础到巅峰(Java 最优解)
java·开发语言·人工智能·python·算法·ai·指针