//AVTree.h #ifndef MY_AVLTREE_H #define MY_AVLTREE_H typedef int ElementType; struct TreeNode { ElementType data; struct TreeNode *left; struct TreeNode *right; int height; }; typedef struct TreeNode TreeNode; typedef TreeNode *Tree; Tree Find(Elemen

二叉搜索树(Binary Search Tree),又名二叉查找树.二叉排序树,是一种简单的二叉树.它的特点是每一个结点的左(右)子树各结点的元素一定小于(大于)该结点的元素.将该树用于查找时,由于二叉树的性质,查找操作的时间复杂度可以由线性降低到O(logN). 当然,这一复杂度只是描述了平均的情况,事实上,具体到每一棵二叉搜索树,查找操作的复杂度与树本身的结构有关.如果二叉树的结点全部偏向一个方向,那么与线性查找将毫无区别.这就牵扯到二叉树的平衡问题,暂时不做考虑. 下面给出二叉搜索树的实现

二叉树(Binary Tree)是最简单的树形数据结构,然而却十分精妙.其衍生出各种算法,以致于占据了数据结构的半壁江山.STL中大名顶顶的关联容器--集合(set).映射(map)便是使用二叉树实现.由于篇幅有限,此处仅作一般介绍(如果想要完全了解二叉树以及其衍生出的各种算法,恐怕要写8~10篇). 1)二叉树(Binary Tree) 顾名思义,就是一个节点分出两个节点,称其为左右子节点:每个子节点又可以分出两个子节点,这样递归分叉,其形状很像一颗倒着的树.二叉树限制了每个节点最多有两个子节

二叉树的一个重要应用就是查找. 二叉搜索树 满足如下的性质: 左子树的关键字 < 节点的关键字 < 右子树的关键字 1. Find(x) 有了上述的性质后,我们就可以像二分查找那样查找给定的关键字x 具体如下: if x < node->val, Search in left sub-tree; else if x > node->val, Search in right sub-tree; else, found it! 2. Insert(x) 插入操作像Find(

Pat1043代码 题目描写叙述: 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 node

1.二叉搜索树:去一个陌生的城市问路到目的地: for each node, all elements in its left subtree are less-or-equal to the node (<=), and all the elements in its right subtree are greater than the node (>). 给予一个已经排序好的整数数组, 生成一个相对合理的二叉搜索树.(?相对合理的?) 给予一个二叉树的根节点,验证该树是否是二叉树搜索树,(

1. 题目 1.1 英文题目 Given an integer array nums where the elements are sorted in ascending order, convert it to a height-balanced binary search tree. A height-balanced binary tree is a binary tree in which the depth of the two subtrees of every node never

题目链接:Recover Binary Search Tree | LeetCode OJ Two elements of a binary search tree (BST) are swapped by mistake. Recover the tree without changing its structure. Note: A solution using O(n) space is pretty straight forward. Could you devise a constan

题目链接:Validate Binary Search Tree | LeetCode OJ Given a binary tree, determine if it is a valid binary search tree (BST). Assume a BST is defined as follows: The left subtree of a node contains only nodes with keys less than the node's key. The right

Sept. 22, 2015 学一道算法题, 经常回顾一下. 第二次重温, 决定增加一些图片, 帮助自己记忆. 在网上找他人的资料, 不如自己动手. 把从底向上树的算法搞通俗一些. 先做一个例子: 9/22/2015 Go over one example to build some muscle memory about this bottom up, O(1) solution to find the root node in subtree function. Sorted List: 1

Given a non-empty binary search tree and a target value, find k values in the BST that are closest to the target. Note: Given target value is a floating point. You may assume k is always valid, that is: k ≤ total nodes. You are guaranteed to have onl

Given a non-empty binary search tree and a target value, find the value in the BST that is closest to the target. Note: Given target value is a floating point. You are guaranteed to have only one unique value in the BST that is closest to the target.

Given an array of numbers, verify whether it is the correct preorder traversal sequence of a binary search tree. You may assume each number in the sequence is unique. Follow up: Could you do it using only constant space complexity? 这道题让给了我们一个一维数组,让我们

Given a binary search tree (BST), find the lowest common ancestor (LCA) of two given nodes in the BST. According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes v and w as the lowest node in T that has

Implement an iterator over a binary search tree (BST). Your iterator will be initialized with the root node of a BST. Calling next() will return the next smallest number in the BST. Note: next() and hasNext() should run in average O(1) time and uses

Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST. 这道题是要求把有序链表转为二叉搜索树,和之前那道Convert Sorted Array to Binary Search Tree 将有序数组转为二叉搜索树思路完全一样,只不过是操作的数据类型有所差别,一个是数组,一个是链表.数组方便就方便在可以通过index直接访问任意一个元

Given an array where elements are sorted in ascending order, convert it to a height balanced BST. 这道题是要将有序数组转为二叉搜索树,所谓二叉搜索树,是一种始终满足左<根<右的特性,如果将二叉搜索树按中序遍历的话,得到的就是一个有序数组了.那么反过来,我们可以得知,根节点应该是有序数组的中间点,从中间点分开为左右两个有序数组,在分别找出其中间点作为原中间点的左右两个子节点,这不就是是二分查找法的核

Two elements of a binary search tree (BST) are swapped by mistake. Recover the tree without changing its structure. Note:A solution using O(n) space is pretty straight forward. Could you devise a constant space solution? confused what "{1,#,2,3}"

Given a binary tree, determine if it is a valid binary search tree (BST). Assume a BST is defined as follows: 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

Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST. Top down 的解题方法: 1. 将LinkedList的值保存到一个数组中,转化成Convert Sorted Array to Binary Search Tree 来解决 时间复杂度为O(n), 空间复杂度为O(n) /** * Definition for singl

验证一个list是不是一个BST的preorder traversal sequence. Given an array of numbers, verify whether it is the correct preorder traversal sequence of a binary search tree. You may assume each number in the sequence is unique. Follow up:Could you do it using only