bst

官网 1043. Is It a Binary Search Tree (25) 时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue A Binary Search Tree (BST) is recursively defined as a binary tree which has the following properties: 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 or equal to the node’s key. Both the left and right subtrees must also be binary search trees. If we swap the left and right subtrees of every node, then the resulting tree is called the Mirror Image of a BST. Now given a sequence of

题目： A Binary Search Tree (BST) is recursively defined as a binary tree which has the following properties: 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 or equal to the node’s key. Both the left and right subtrees must also be binary search trees. If we swap the left and right subtrees of every node, then the resulting tree is called the Mirror Image of a BST. Now given a sequence of integer keys, you are supposed to tell if it is the preorder traversal sequence of a BST or the mirror

1043 Is It a Binary Search Tree （25 分） 原文链接： http://kakazai.cn/index.php/Kaka/Pat/query/id/133 文章目录 题目 题意分析 知识点与坑点 一、定义法 算法思路 代码-c++版 代码-python版 题目 题目链接： https://pintia.cn/problem-sets/994805342720868352/problems/994805440976633856 A Binary Search Tree (BST) is recursively defined as a binary tree which has the following properties: 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 or equal to the node’s key. Both the left and right subtrees must also be binary search trees. If we swap the

ECE241 PROJECT 1: Sorting and Searching Due: October 24, 2019, 11PM on gradescope Introduction: In today’s Internet-dominated world we take for granted the ability of computers to search through vast quantities of data to find information that is of interest to us. The ability of search engines to find data quickly is the result of many programmers’ and algorithm designers’ efforts over a number of years. In Project 1, you will have the opportunity to learn the basics to become a successful future programmer by implementing algorithms for the Million Song Dataset, which will both sort and

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: When implementing a Hashtable using an array, we inherit the constant time indexing of the array. What are the reasons for implementing a Hashtable with a Binary Search Tree since it offers search with O(logn)? Why not just use a Binary Search Tree directly? 回答1: If the elements don't have a total order (i.e. the "greater than" and "less than" is not be defined for all pairs or it is not consistent between elements), you can't compare all pairs, thus you can't use a BST directly, but nothing's stopping you from indexing the BST by the hash

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: To construct a BST from the preorder traversal given, if I try to insert in the BST in the same order as given in preorder, I get the BST. So, we don't to create the in-order by sorting of the elements or perform any other alogrithm? Is there an example which shows that tree can't be constructed by just inserting the elements ? 回答1: You're correct... you can just insert the nodes in the order given by a preorder traversal to rebuild the tree. The first node inserted must be placed at the right place, since it's the root and a preorder

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 由 翻译 强力驱动 问题: Reference: I was asked this question @MS SDE interview, 3rd round. And it's not a homework problem. I also gave it a thought and mentioning my approach below. Question: Modify a BST so that it becomes as balanced as possible. Needless to mention, you should do it as efficient as possible. Hint: Interviewer said that this is a logical question, if you think differently you will get the answer. No difficult coding involved. --> Having said that, I do not think he was expecting me to point to AVL/RB Trees. My Solution: I proposed that

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 由 翻译 强力驱动 问题: I am trying to convert a unix epoch timestamps to a date-time object using as.POSIXct() I need to specify timezones (either Europe/London or UTC) when I call as.POSIXct(). If I run > t <- as . POSIXct ( 1445329330 , tz = "Europe/London" , origin = "1970-01-01" ) > t R returns "2015-10-20 09:22:10 BST" Warning messages: 1: In as.POSIXlt.POSIXct(x, tz) : unknown timezone 'default/Europe/London' 2: In as.POSIXlt.POSIXct(x, tz) : unknown timezone 'default/Europe/London' I have tried specifying tz="BST", but this also returns warnings

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: I know there are ways to construct a tree from pre-order traversal (as an array). The more common question is to construct it, given the inorder and pre-order traversals. In this case, although the inorder traversal is redundant, it definitely makes things easier. Can anybody give me an idea how to do it for a post-order traversal? Both iterative and recursive solutions are required. I tried to do it iteratively using stack, but couldn't at all get the logic right, so got a horrible messy tree. Same went for recursion. 回答1: If you have an

关于二叉查找树的操作还是挺重要的，所以在此实现了BST的相关操作。 二叉查找树（Binary Search Tree），（又：二叉搜索树，二叉排序树）它或者是一棵空树，或者是具有下列性质的二叉树： 若它的左子树不空，则左子树上所有结点的值均小于它的根结点的值； 若它的右子树不空，则右子树上所有结点的值均大于它的根结点的值； 它的左、右子树也分别为二叉排序树。 首先实现BST的遍历操作吧，一般面试也喜欢问，比较重要的是各个遍历的非递归操作，在这里也就只实现非递归操作； 其中先序和中序遍历的非递归操作比较相似，主要思想就是将每个节点当做一个根节点进行操作。后序遍历稍微不同。下面进行代码的演示，以及相关解释： 首先针对先序遍历 Stack<TreeNode> snode = new Stack<>(); TreeNode p = root; while (!snode.isEmpty() || p != null) { if (p!=null) { System.out.print(p.value+" "); snode.push(p); p = p.left; }else{ TreeNode q = snode.pop(); p = q.right; } } 可以看到代码中利用了一个栈进行辅助，因为先序遍历的顺序是根-左-右，所以在向左节点进行遍历的时候，将节点输出，同时将节点放入栈中