Hi, I'm Louislan ! 👋🏼

题目地址 https://leetcode.com/problems/convert-sorted-array-to-binary-search-tree/ 题目描述 Given an array where elements are sorted in ascending order, convert it to a height balanced BST. For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1. Example: Given the sorted array: [-10,-3,0,5,9], One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST: 0 / \ -3 9 / /....

题目地址 https://leetcode.com/problems/binary-tree-level-order-traversal-ii/ 题目描述 Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root). For example: Given binary tree {3,9,20,#,#,15,7}, 3 / \ 9 20 / \ 15 7 return its bottom-up level order traversal as: [ [15,7], [9,20], [3] ] 思路 从底部层序遍历其实还是从顶部开始遍历，只不过最后存储的方式有所改变，可以参见我之前的博文 Binary Tree Level Order Traversal 关键点解析 代码 解法一： class Solution { public: vector<v....

题目地址 https://leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/ 题目描述 Given inorder and postorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. 思路 这道题要求从中序和后序遍历的结果来重建原二叉树，我们知道中序的遍历顺序是左-根-右，后序的顺序是左-右-根，对于这种树的重建一般都是采用递归来做，可参见我之前的一篇博客Convert Sorted Array to Binary Search Tree 将有序数组转为二叉搜索树。针对这道题，由于后序的顺序的最后一个肯定是根，所以原二叉树的根节点可以知道，题目中给了一个很关键的条件就是树中没有相同元素，有了这个条件我们就可以在中序遍历中也定位出根节点的位置，并以根节点的位置将中序遍历拆分为左右两个部分，分别对其递....

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定制皮肤 前景提示：本人是个纯后端接口开发(java)，对于前端涉略较浅，这篇文章的目的是为了方便那些同道中人或者前端小白能够基于现有的solo皮肤简单开(bai)发(piao)出一套适合自己的博客样式，其间若有冒犯，欢迎来犯，一起进步！

题目地址 https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/ 题目描述 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 思路 这道题要求用先序和中序遍历来建立二叉树，跟之前那道 Construct Binary Tree from Inorder and Postorder Traversal 原理基本相同，针对这道题，由于先序的顺序的第一个肯定是根，所以原二叉树的根节点可以知道，题目中给了一....

题目地址 https://leetcode.com/problems/maximum-depth-of-binary-tree/ 题目描述 Given a binary tree, find its maximum depth. The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node. Note: A leaf is a node with no children. Example: Given binary tree [3,9,20,null,null,15,7], 3 / \ 9 20 / \ 15 7 return its depth = 3. 思路 求二叉树的最大深度问题用到深度优先搜索 Depth First Search，递归的完美应用，跟求二叉树的最小深度问题原理相同，参见代码如下： C++ 解法一： class Solution { public: int maxDepth(TreeNode* r....

题目地址 https://leetcode.com/problems/binary-tree-zigzag-level-order-traversal/ 题目描述 Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between). For example: Given binary tree [3,9,20,null,null,15,7], 3 / \ 9 20 / \ 15 7 return its zigzag level order traversal as: [ [3], [20,9], [15,7] ] 思路 这道二叉树的之字形层序遍历是之前那道 Binary Tree Level Order Traversal 的变形，不同之处在于一行是从左到右遍历，下一行是从右往左遍历，交叉往返的之字形的层序遍历。....

题目地址 https://leetcode.com/problems/binary-tree-level-order-traversal/ 题目描述 Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level). For example: Given binary tree {3,9,20,#,#,15,7}, 3 / \ 9 20 / \ 15 7 return its level order traversal as: [ [3], [9,20], [15,7] ] 思路 层序遍历二叉树是典型的广度优先搜索 BFS 的应用，但是这里稍微复杂一点的是，要把各个层的数分开，存到一个二维向量里面，大体思路还是基本相同的，建立一个 queue，然后先把根节点放进去，这时候找根节点的左右两个子节点，这时候去掉根节点，此时 queue 里的元素就是下一层的所有节点，用一个 for 循环遍历它们，然后存到一个一维向量里，....