Leetcode 0199. Binary Tree Right Side View
199. Binary Tree Right Side ViewGiven the root of a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom. Example 1: Input: root = [1,2,3,null,5,null,4] Output: [1,3,4] Explanation: Example 2: Input: root = [1,2,3,4,null,null,null,5] Output: [1,3,4,5] Explanation: Example 3: Input: root = [1,null,3] Output: [1,3] Example 4: Input: root = [] Output: [] 题目大意给定一棵二叉树的根节点 root，想象自己站在树的右侧，返回从顶部到底...
Leetcode 0189. 轮转数组
189. 轮转数组给定一个整数数组 nums，将数组中的元素向右轮转 k 个位置，其中 k 是非负数。 示例 1: 123456输入: nums = [1,2,3,4,5,6,7], k = 3输出: [5,6,7,1,2,3,4]解释:向右轮转 1 步: [7,1,2,3,4,5,6]向右轮转 2 步: [6,7,1,2,3,4,5]向右轮转 3 步: [5,6,7,1,2,3,4] 示例 2: 12345输入：nums = [-1,-100,3,99], k = 2输出：[3,99,-1,-100]解释: 向右轮转 1 步: [99,-1,-100,3]向右轮转 2 步: [3,99,-1,-100] 题目大意给定一个整数数组 nums 和非负整数 k，将数组元素向右轮转 k 个位置（即每个元素向右移动 k 位，末尾元素移动到开头）。要求尽可能优化时间和空间复杂度。 核心解题思路：三次反转法常规思路（如临时数组、多次右移）要么空间复杂度高（O (n)），要么时间复杂度高（O (nk)）。三次反转法通过巧妙的反转操作，实现 O (n) 时间复杂度 + O (1) 空间复杂度...
Leetcode 0180. 连续出现的数字
180. 连续出现的数字表：Logs 12345678+-------------+---------+| Column Name | Type |+-------------+---------+| id | int || num | varchar |+-------------+---------+在 SQL 中，id 是该表的主键。id 是一个自增列。 找出所有至少连续出现三次的数字。 返回的结果表中的数据可以按 任意顺序 排列。 结果格式如下面的例子所示： 示例 1: 123456789101112131415161718192021输入：Logs 表：+----+-----+| id | num |+----+-----+| 1 | 1 || 2 | 1 || 3 | 1 || 4 | 2 || 5 | 1 || 6 | 2 || 7 | 2 |+----+-----+输出：Result 表：+-----------------+| ConsecutiveNums |+----...
Leetcode 0162. Find Peak Element
162. Find Peak Element题目A peak element is an element that is greater than its neighbors. Given an input array nums, where nums[i] ≠ nums[i+1], find a peak element and return its index. The array may contain multiple peaks, in that case return the index to any one of the peaks is fine. You may imagine that nums[-1] = nums[n] = -∞. Example 1: Input: nums = [1,2,3,1] Output: 2 Explanation: 3 is a peak element and your function should return the index number 2. Example 2: Input: nums = [1,2,1,3,...
Leetcode 0155. Min Stack
155. Min StackDesign a stack that supports push, pop, top, and retrieving the minimum element in constant time. Implement the MinStack class: MinStack() initializes the stack object. void push(int val) pushes the element val onto the stack. void pop() removes the element on the top of the stack. int top() gets the top element of the stack. int getMin() retrieves the minimum element in the stack. You must implement a solution with O(1) time complexity for each function. 题目大意设计一个支持 push（压栈）、p...
Leetcode 0151. Reverse Words in a String
151. Reverse Words in a StringGiven an input string s, reverse the order of the words. A word is defined as a sequence of non-space characters. The words in s will be separated by at least one space. Return a string of the words in reverse order concatenated by a single space. Note that s may contain leading or trailing spaces or multiple spaces between two words. The returned string should only have a single space separating the words. Do not include any extra spaces. Example 1: 12Input: s =...
Leetcode 0150. Evaluate Reverse Polish Notation
150. Evaluate Reverse Polish NotationYou are given an array of strings tokens that represents an arithmetic expression in a Reverse Polish Notation. Evaluate the expression. Return an integer that represents the value of the expression. Note that: The valid operators are '+', '-', '*', and '/'. Each operand may be an integer or another expression. The division between two integers always truncates toward zero. There will not be any division by zero. The input ...
Leetcode 0148. 排序链表
148. 排序链表给你链表的头结点 head ，请将其按 升序 排列并返回 排序后的链表 。 示例 1： 12输入：head = [4,2,1,3]输出：[1,2,3,4] 示例 2： 12输入：head = [-1,5,3,4,0]输出：[-1,0,3,4,5] 示例 3： 12输入：head = []输出：[] 题目大意给定链表的头节点 head，要求将链表按升序排列并返回排序后的链表头节点。 解题思路对于链表排序，最适合的高效算法是归并排序，原因如下： 归并排序的时间复杂度为 O (n log n)，是链表排序的最优选择 链表的归并操作不需要像数组那样额外分配 O (n) 的空间 链表的中点查找可以通过快慢指针高效实现 归并排序的核心步骤： 分解：使快慢指针找到链表中点，将链表分成两部分 递归：对左右两部分分别进行排序 合并：将两个已排序的链表合并成一个有序链表 1234567891011121314151617181920212223242526272829303132333435363738394041424344454647484950515253...
Leetcode 0146. LRU Cache
146. LRU CacheDesign a data structure that follows the constraints of a Least Recently Used (LRU) cache. Implement the LRUCache class: LRUCache(int capacity) Initialize the LRU cache with positive size capacity. int get(int key) Return the value of the key if the key exists, otherwise return -1. void put(int key, int value) Update the value of the key if the key exists. Otherwise, add the key-value pair to the cache. If the number of keys exceeds the capacity from this operation, evict the l...
Leetcode 0145. Binary Tree Postorder Traversal
145. Binary Tree Postorder TraversalGiven the root of a binary tree, return the postorder traversal of its nodes' values. Example 1: Input: root = [1,null,2,3] Output: [3,2,1] Explanation: Example 2: Input: root = [1,2,3,4,5,null,8,null,null,6,7,9] Output: [4,6,7,5,2,9,8,3,1] Explanation: Example 3: Input: root = [] Output: [] Example 4: Input: root = [1] Output: [1] 题目大意给定一棵二叉树的根节点 root，返回其节点值的后序遍历结果。后序遍历的顺序是「左子树 → 右子树 → 根节点」，遵循 "左 - 右 - 根" 的递归逻辑。 解题思路后序遍...
Leetcode 0144. Binary Tree Preorder Traversal
144. Binary Tree Preorder TraversalGiven the root of a binary tree, return the preorder traversal of its nodes' values. Example 1: Input: root = [1,null,2,3] Output: [1,2,3] Explanation: Example 2: Input: root = [1,2,3,4,5,null,8,null,null,6,7,9] Output: [1,2,4,5,6,7,3,8,9] Explanation: Example 3: Input: root = [] Output: [] Example 4: Input: root = [1] Output: [1] 题目大意给定一棵二叉树的根节点 root，返回其节点值的前序遍历结果。前序遍历的顺序是「根节点 → 左子树 → 右子树」，遵循 "根 - 左 - 右" 的递归逻辑。 解题思路前序遍历的...
Leetcode 0142. Linked List Cycle II
142. Linked List Cycle IIGiven the head of a linked list, return the node where the cycle begins. If there is no cycle, return null. There is a cycle in a linked list if there is some node in the list that can be reached again by continuously following the next pointer. Internally, pos is used to denote the index of the node that tail's next pointer is connected to (0-indexed). It is -1 if there is no cycle. Note that pos is not passed as a parameter. Do not modify the linked list. Examp...
Leetcode 0131. Palindrome Partitioning
131. Palindrome PartitioningGiven a string s, partition s such that every substring of the partition is a palindrome. Return all possible palindrome partitioning of s. Example 1: 12Input: s = "aab"Output: [["a","a","b"],["aa","b"]] Example 2: 12Input: s = "a"Output: [["a"]] 题目大意给定一个字符串 s，将其分割成若干个子串，使得每个子串都是回文串。返回所有可能的分割方案。 例如： 输入 s = "aab"，输出 [["a","a","b"],["aa&quo...
Leetcode 0116. Populating Next Right Pointers in Each Node
116. Populating Next Right Pointers in Each NodeYou are given a perfect binary tree where all leaves are on the same level, and every parent has two children. The binary tree has the following definition: 123456struct Node { int val; Node *left; Node *right; Node *next;} Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL. Initially, all next pointers are set to NULL. Example 1: 123Input: root = [1...
Leetcode 0112. Path Sum
112. Path SumGiven the root of a binary tree and an integer targetSum, return true if the tree has a root-to-leaf path such that adding up all the values along the path equals targetSum. A leaf is a node with no children. Example 1: 123Input: root = [5,4,8,11,null,13,4,7,2,null,null,null,1], targetSum = 22Output: trueExplanation: The root-to-leaf path with the target sum is shown. Example 2: 123456Input: root = [1,2,3], targetSum = 5Output: falseExplanation: There are two root-to-leaf pat...
Leetcode 0110. Balanced Binary Tree
110. Balanced Binary TreeGiven a binary tree, determine if it is height-balanced. Example 1: 12Input: root = [3,9,20,null,null,15,7]Output: true Example 2: 12Input: root = [1,2,2,3,3,null,null,4,4]Output: false Example 3: 12Input: root = []Output: true 题目大意给定一棵二叉树，判断它是否是高度平衡的。高度平衡平衡二叉树 ** 的定义是：二叉树的每个节点的左右两个子树的高度差的绝对值不超过 1。 例如： 输入二叉树 [3,9,20,null,null,15,7]，每个节点的左右子树高度差均不超过 1，返回 true； 输入二叉树 [1,2,2,3,3,null,null,4,4]，根节点的左子树高度为 3，右子树高度为 1，差为 2，返回 false。 解题思路判断平衡二叉树的核心是计算每个节点的左右子树高度，并检查高...
Leetcode 0108. Convert Sorted Array to Binary Search Tree
108. Convert Sorted Array to Binary Search TreeGiven an integer array nums where the elements are sorted in ascending order, convert it to a *height-balanced* binary search tree. Example 1: 123Input: nums = [-10,-3,0,5,9]Output: [0,-3,9,-10,null,5]Explanation: [0,-10,5,null,-3,null,9] is also accepted: Example 2: 123Input: nums = [1,3]Output: [3,1]Explanation: [1,null,3] and [3,1] are both height-balanced BSTs. 题目大意给定一个升序排列的整数数组 nums，将其转换为一棵高度平衡的二叉搜索树（BST）。高度平衡的定义是：二叉树的左右两个子树的高度差的绝对值不超过 ...
Leetcode 0107. Binary Tree Level Order Traversal II
107. Binary Tree Level Order Traversal IIGiven the root of a binary tree, return the bottom-up level order traversal of its nodes' values. (i.e., from left to right, level by level from leaf to root). Example 1: 12Input: root = [3,9,20,null,null,15,7]Output: [[15,7],[9,20],[3]] Example 2: 12Input: root = [1]Output: [[1]] Example 3: 12Input: root = []Output: [] 题目大意给定一棵二叉树的根节点 root，返回其节点值的自底向上的层序遍历结果。即按「从叶子节点所在层到根节点所在层」的顺序逐层返回，每层内部仍保持「从左到右」的顺序。 例如： 输入二叉树 [3,9,20,null,null,15,7]，其自底向上...
Leetcode 0106. Construct Binary Tree from Inorder and Postorder Traversal
106. Construct Binary Tree from Inorder and Postorder TraversalGiven two integer arrays inorder and postorder where inorder is the inorder traversal of a binary tree and postorder is the postorder traversal of the same tree, construct and return the binary tree. Example 1: 12Input: inorder = [9,3,15,20,7], postorder = [9,15,7,20,3]Output: [3,9,20,null,null,15,7] Example 2: 12Input: inorder = [-1], postorder = [-1]Output: [-1] 题目大意给定一棵二叉树的中序遍历数组 inorder 和后序遍历数组 postorder，请构造并返回这棵二叉树。 例如： ...
Leetcode 0105. Construct Binary Tree from Preorder and Inorder Traversal
105. Construct Binary Tree from Preorder and Inorder Traversal题目Given preorder and inorder traversal of a tree, construct the binary tree. **Note:**You may assume that duplicates do not exist in the tree. For example, given preorder = [3,9,20,15,7] inorder = [9,3,15,20,7] Return the following binary tree: 3 / \ 9 20 / \ 15 7 题目大意根据一棵树的前序遍历与中序遍历构造二叉树。 注意:你可以假设树中没有重复的元素。 解题思路 inorder_map：用于快速查找中序遍历中元素对应的索引，时间复杂度 O (1) pre：保存前序遍历数组的副本，避免在递归中反复传递参数 确定根节点：前序遍历的第一个元素为当前树的根节点 划分左右...
Leetcode 0104. Maximum Depth of Binary Tree
104. Maximum Depth of Binary TreeGiven the root of a binary tree, return its maximum depth. A binary tree's maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node. Example 1: 12Input: root = [3,9,20,null,null,15,7]Output: 3 Example 2: 12Input: root = [1,null,2]Output: 2 题目大意给定一棵二叉树的根节点 root，返回该二叉树的最大深度。二叉树的最大深度是指从根节点到最远叶子节点的最长路径上的节点数量。 例如： 输入二叉树 [3,9,20,null,null,15,7]，其最大深度为 3（路径：3 → 20 → 15 或 3 → 20 → 7，均包含 3 个节点）。 解题思路计算二叉树的最...
Leetcode 0103. Binary Tree Zigzag Level Order Traversal
103. Binary Tree Zigzag Level Order Traversal Given the root of a binary tree, return the zigzag level order traversal of its nodes' values. (i.e., from left to right, then right to left for the next level and alternate between). Example 1: 12Input: root = [3,9,20,null,null,15,7]Output: [[3],[20,9],[15,7]] Example 2: 12Input: root = [1]Output: [[1]] Example 3: 12Input: root = []Output: [] 题目大意给定一棵二叉树的根节点，返回其节点值的锯齿形层序遍历。即：第一层从左到右，第二层从右到左，第三层再从左到右，以此类推，交替进行。 例如： 输入 root = [3,9,20,null...
Leetcode 0102. Binary Tree Level Order Traversal
102. Binary Tree Level Order TraversalGiven the root of a binary tree, return the level order traversal of its nodes' values. (i.e., from left to right, level by level). Example 1: 12Input: root = [3,9,20,null,null,15,7]Output: [[3],[9,20],[15,7]] Example 2: 12Input: root = [1]Output: [[1]] Example 3: 12Input: root = []Output: [] 题目大意给定一棵二叉树的根节点 root，返回其节点值的层序遍历结果（即从左到右、逐层遍历）。结果需以二维数组形式呈现，每一层的节点值构成一个子数组（例如，第一层 [根节点]，第二层 [左子节点, 右子节点]，以此类推）。 解题思路层序遍历的核心是按照树的层次依次访问节点，从根节点开始，先访问第一层（根节点）...
Leetcode 0101. Symmetric Tree
101. Symmetric TreeGiven the root of a binary tree, check whether it is a mirror of itself (i.e., symmetric around its center). Example 1: 12Input: root = [1,2,2,3,4,4,3]Output: true Example 2: 12Input: root = [1,2,2,null,3,null,3]Output: false 题目大意给定一棵二叉树的根节点 root，判断该二叉树是否是对称的（即围绕中心轴镜像对称）。 例如： 输入二叉树 [1,2,2,3,4,4,3]，其左子树与右子树成镜像，故返回 true； 输入二叉树 [1,2,2,null,3,null,3]，左子树与右子树不镜像，故返回 false。 解题思路判断二叉树是否对称的核心是比较左子树与右子树是否成镜像，即： 左子树的左节点需与右子树的右节点值相等； 左子树的右节点需与右子树的左节点值相等。 1. 递归法设计一个辅助函数，比较两个子树...
Leetcode 0100. Same Tree
100. Same TreeGiven the roots of two binary trees p and q, write a function to check if they are the same or not. Two binary trees are considered the same if they are structurally identical, and the nodes have the same value. Example 1: 12Input: p = [1,2,3], q = [1,2,3]Output: true Example 2: 12Input: p = [1,2], q = [1,null,2]Output: false Example 3: 12Input: p = [1,2,1], q = [1,1,2]Output: false 题目大意给定两棵二叉树的根节点 p 和 q，判断这两棵树是否相同。两棵树相同的定义是：结构完全相同，且对应节点的值也相同。 例如： 输入两棵结构和节点值均相同的树 [1,2,3]...
