A Tree

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★ ➤微信公众号：山青咏芝（shanqingyongzhi） ➤博客园地址：山青咏芝（ https://www.cnblogs.com/strengthen/ ） ➤GitHub地址： https://github.com/strengthen/LeetCode ➤原文地址： https://www.cnblogs.com/strengthen/p/10693873.html ➤如果链接不是山青咏芝的博客园地址，则可能是爬取作者的文章。 ➤原文已修改更新！强烈建议点击原文地址阅读！支持作者！支持原创！ ★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★ Serialization is the process of converting a data structure or object into a sequence of bits so that it can be stored in a file or memory buffer, or transmitted across a network connection link to be reconstructed later in the same or another computer environment.

Trees are fundamental in many branches of computer science (Pun definitely intended). Current stateof-the art parallel computers such as Thinking Machines’ CM-5 are based on fat trees . Quad- and octal-trees are fundamental to many algorithms in computer graphics. This problem involves building and traversing binary trees. Given a sequence of binary trees, you are to write a program that prints a level-order traversal of each tree. In this problem each node of a binary tree contains a positive integer and all binary trees have have fewer than 256 nodes. In a level-order traversal of a tree,

Problem LeetCode Given the root of a tree, you are asked to find the most frequent subtree sum. The subtree sum of a node is defined as the sum of all the node values formed by the subtree rooted at that node (including the node itself). So what is the most frequent subtree sum value? If there is a tie, return all the values with the highest frequency in any order. Examples 1 Input: 5 / \ 2 -3 return [2, -3, 4], since all the values happen only once, return all of them in any order. Examples 2 Input: 5 / \ 2 -5 return [2], since 2 happens twice, however -5 only occur once. Note: You may assume

递归确实是一种优雅强大的技术， 但是好多代码库都偏爱使用迭代，即使使用递归， 也都往往对递归调用的最大栈深度提前做预估或限制等。可能考虑递归性能一般低于迭代。有些问题，我们可能先是使用递归解决， 然后再转变成对应的迭代版本， 练习递归到迭代的转换，也有助于我们理解问题的递归结构。 树是典型的递归定义数据结构, 对应的操作也是递归的实现，如二叉树的遍历: type node struct { link [2]*node data rune } func preOrder(root *node) { if root != nil { fmt.Printf("%c ", root.data) preOrder(root.link[0]) //left subtree preOrder(root.link[1]) //right subtree } } func inOrder(root *node) { if root != nil { inOrder(root.link[0]) //left subtree fmt.Printf("%c ", root.data) inOrder(root.link[1]) //right subtree } } func postOrder(root *node) { if root != nil { postOrder(root.link[0])

Problem LeetCode Given the root of a Binary Search Tree (BST), convert it to a Greater Tree such that every key of the original BST is changed to the original key plus sum of all keys greater than the original key in BST. As a reminder, a binary search tree is a tree that satisfies these constraints: The left subtree of a node contains only nodes with keys less than the node's key. The right subtree of a node contains only nodes with keys greater than the node's key. Both the left and right subtrees must also be binary search trees. Note: This question is the same as 538: https://leetcode.com

Problem LeetCode Given the root of a Binary Search Tree (BST), convert it to a Greater Tree such that every key of the original BST is changed to the original key plus sum of all keys greater than the original key in BST. As a reminder, a binary search tree is a tree that satisfies these constraints: The left subtree of a node contains only nodes with keys less than the node's key. The right subtree of a node contains only nodes with keys greater than the node's key. Both the left and right subtrees must also be binary search trees. Note: This question is the same as 1038: https://leetcode.com

<br> Given a tree, rearrange the tree in in-order so that the leftmost node in the tree is now the root of the tree, and every node has no left child and only 1 right child. ``` Example 1: Input: [5,3,6,2,4,null,8,1,null,null,null,7,9] 5 / \ 3 6 / \ 2 4 8 / / 1 7 9 Output: [1,null,2,null,3,null,4,null,5,null,6,null,7,null,8,null,9] 1 2 3 4 5 6 7 8 9 Note: 1. The number of nodes in the given tree will be between 1 and 100. 2. Each node will have a unique integer value from 0 to 1000. <br> 这道题给了一棵二叉树，让我们对其进行重排序，使得最左结点变为根结点，而且整个树不能有左子结点，如题目中的例子所示，排序后的结果是一条向右下方延伸的直线。如果我们仔细观察题目中的例子，可以发现遍历顺序其实是 左->根

[TOC] 本文参考自 梁晏成《树上数据结构》 ，感谢他在雅礼集训的讲解。 转化成序列问题 dfs序 按照 $dfs$ 的入栈顺序形成一个序列。 例如对于这棵树 它的 $dfs$ 序就是 $1~2~3~4~5~6~7~8$ 。（假设我遍历儿子是从左到右的） 树链剖分的运用 对于这个我们常常配合 树链剖分 来使用。 这样对于一个点，它的子树编号是连续的一段区间，便于做子树修改以及查询问题。 重链上所有节点的标号也是连续的一段区间。 所以我们可以解决大部分链或子树修改以及查询的问题，十分的优秀。 也就是常常把树上问题转化成序列问题的通用解法。 括号序列 $dfs$ 时候，某个节点入栈时加入左括号，出栈时加入右括号。 也就是在 $dfs$ 序旁边添加括号。 同样对于上面那颗树 。 为了方便观看，我们在其中添入一些数字。 它的括号序列就是 $(1(2)(3(4)(5(6(7))))(8))$ 。 求解树上距离问题 这个可以对于一些有关于树上距离的问题有用，比如 BZOJ1095 [ZJOI2007] Hide 捉迷藏 (括号序列 + 线段树) 也就是对于树上两点的距离，就是他们中间未匹配的括号数量。这个是很显然的，因为匹配的括号必定不存在于他们之间的路径上，其他的都存在于他们的路径上。 也就是说向上路径的括号是 $)$ 向下路径的括号就是 $($ 。 树上莫队转化成普通莫队 令 $L_x$

众所周知 ，莫队是由莫涛大神提出的，一种 玄学毒瘤暴力骗分 区间操作算法，它以简短的框架、 简单易记的板子 和优秀的复杂度闻名于世。然而由于莫队算法应用的毒瘤，很多可做的莫队模板题都有着较高的难度评级，令很多初学者望而却步。然而，如果你真正理解了莫队的算法原理，那么它用起来还是很简单的。 当然某些套左套右的毒瘤除外 0、前置芝士： 莫队算法还是比较独立的。不过你还是得了解了解以下的一些知识： \(1\) 、分块的基本思想（开根号等） \(2\) 、STL中 sort 的用法（手写cmp函数或重载运算符实现结构体的多关键字排序） \(3\) 、基(du)础(liu)的卡常技巧（包含 #pragma GCC optimize 系列） \(4*\) 、倍增/树剖 求LCA（树上莫队所需） \(5*\) 、数值离散化（用于应付很多题目） 至此全部完毕。撒花~~（雾 诶，别走啊qwq，我可不是在劝退qwq，如果你认为自己不懂这些东西也没关系，往下看吧qwq 1、莫队算法是个啥 来历： 前面已经介绍过了（逃 有兴趣的同学可以看一下 莫涛大神的知乎 然而这个算法到底是用来搞什么操作的呢？我们先看个例题： Luogu P1972 [SDOI2009]HH的项链 题目描述 HH 有一串由各种漂亮的贝壳组成的项链。HH 相信不同的贝壳会带来好运，所以每次散步完后，他都会随意取出一段贝壳

问题： I've been wondering whether there is a good "git export" solution that creates a copy of a tree without the .git repository directory. 我一直在想是否有一个好的“ git export”解决方案来创建没有 .git 存储库目录的树的副本。 There are at least three methods I know of: 我至少知道三种方法： git clone followed by removing the .git repository directory. git clone 然后删除 .git 存储库目录。 git checkout-index alludes to this functionality but starts with "Just read the desired tree into the index..." which I'm not entirely sure how to do. git checkout-index 暗示了此功能，但以“只需将所需的树读入索引...”开头，我不确定该怎么做。 git-export is a third party script that essentially does