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…

4.7 Design an algorithm and write code to find the first common ancestor of two nodes in a binary tree. Avoid storing additional nodes in a data structure. NOTE: This is not necessarily a binary search tree. LeetCode上的原题,请参见我之前的博客Lowest Common Ancest…

Return the root node of a binary search tree that matches the given preorder traversal. (Recall that a binary search tree is a binary tree where for every node, any descendant of node.left has a value < node.val, and any descendant of node.right has…

1043 Is It a Binary Search Tree(25 分) 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 nod…

既上篇关于二叉搜索树的文章后,这篇文章介绍一种针对二叉树的新的中序遍历方式,它的特点是不需要递归或者使用栈,而是纯粹使用循环的方式,完成中序遍历. 线索二叉树介绍 首先我们引入“线索二叉树”的概念: "A binary tree is threaded by making all right child pointers that would normally be null point to the inorder successor of the node, and all left chi…

[题目] 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…

树的题目,往往可以用到三种遍历.以及递归,因为其结构上天然地可以往深处递归,且判断条件也往往不复杂(左右子树都是空的). LeetCode 98题讲的是,判断一棵树是不是二叉搜索树. 题目中给的是标准定义:即一个二叉搜索树(binary search tree,简称BST)的每个节点都满足:1.其左侧的整个子树的值都比它小:2.其右侧的整个子树的值都比他大:3.它的左右子树依旧保持二叉搜索树的结构. 第0步 一开始理解错了条件,容易理解成,每个节点只需要比它的左子树大,比它的右子树小,于是得到了…

题目就是给出一棵二叉搜索树,已知根节点为0,并且给出一个序列要插入到这课二叉树中,求这棵二叉树层次遍历后的序列. 用结构体建立节点,val表示该节点存储的值,left指向左孩子,right指向右孩子.中序遍历的顺序正好是序列从小到大的顺序,因此中序遍历的时候顺便赋值就可以了,最后层次遍历输出. 思路一:中序遍历的时候赋值 #include <iostream> #include <cstdio> #include <algorithm> #include <str…

1064 Complete Binary Search Tree (30)(30 分) 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…

★实验任务 可怜的 Bibi 刚刚回到家,就发现自己的手机丢了,现在他决定回头去搜索 自己的手机. 现在我们假设 Bibi 的家位于一棵二叉树的根部.在 Bibi 的心中,每个节点 都有一个权值 x,代表他心中预感向这个节点走可能找回自己手机的程度(虽然 他的预感根本不准).当 Bibi 到达一个节点时,如果该节点有未搜索过的儿子节 点,则 Bibi 会走向未搜索过的儿子节点进行搜索,否则就返回父亲节点.如果 某节点拥有两个未搜索过的儿子节点,Bibi 会选择先搜索权值大的儿子节点. 假设 Bi…

144. Binary Tree Preorder Traversal 前序的非递归遍历:用堆来实现 如果把这个代码改成先向堆存储左节点再存储右节点,就变成了每一行从右向左打印 如果用队列替代堆,并且先存储左节点,再存储右节点,就变成了逐行打印 class Solution { public: vector<int> preorderTraversal(TreeNode* root) { vector<int> result; if(root == NULL) return res…

二叉搜索树是常用的概念,它的定义如下: 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…

开一个指针数组,中序遍历这个二叉搜索树,将节点的指针依次保存在数组里, 然后寻找两处逆序的位置, 中序便利里BST得到的是升序序列 ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ 二叉搜索树(BST)中的两个节点不小心被交换了下. 不改变其结构的情况下恢复这个树. 笔记: 用O(n)的空间复杂度的方法很直接.你能否设计一个常量空间的解决法方案? +++++++++++++++++…

题目信息 1064. Complete Binary Search Tree (30) 时间限制100 ms 内存限制65536 kB 代码长度限制16000 B 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 tha…

一.简介 在计算机科学中,二叉树是每个结点最多有两个子树的树结构.通常子树被称作“左子树”(left subtree)和“右子树”(right subtree).二叉树常被用于实现二叉查找树和二叉堆. 一棵深度为k,且有2^k-1个结点的二叉树,称为满二叉树.这种树的特点是每一层上的结点数都是最大结点数.而在一棵二叉树中,除最后一层外,若其余层都是满的,并且或者最后一层是满的,或者是在右边缺少连续若干结点,则此二叉树为完全二叉树.具有n个结点的完全二叉树的深度为floor(log2n)+1.深度…

给定一个二叉搜索树的两个节点,找出他们的最近公共祖先,如, _______6______ / \ ___2__ ___8__ / \ / \ 0 4 7 9 / \ 3 5 2和8的最近公共祖先是6,2和4的最近公共祖先是2,假设找的3和5 TreeNode* l =lowestCommonAncestor(root->left,p,q) ; TreeNode* r =lowestCommonAncestor(root->right,p,q) ; 返回到4时两个都不是Nullptr,那么要返回…

［抄题］: 给一个二叉查找树以及一个节点,求该节点的中序遍历后继,如果没有返回null ［思维问题］: 不知道分合算法和后序节点有什么关系:直接return表达式就行了,它自己会终止的. ［一句话思路］: 比root大时直接扔右边递归,比root小时 考虑是左边递归还是就是root ［输入量］:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入): ［画图］: ［一刷］: 要定义left节点,留着做后续的比较 ［二刷］: ［三刷］: ［四刷］: ［五刷］: ［总结］…

［抄题］: 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 m…

还是中序遍历建树 #include<bits/stdc++.h> using namespace std; ; struct node { int data; int L,R; }s[N]; int in[N]; ; void inorder(int root) { ) return; inorder(s[root].L); s[root].data=in[cnt++]; inorder(s[root].R); } vector<int>vec; void bfs() { queu…

给定一个二叉搜索树,同时给定最小边界L 和最大边界 R.通过修剪二叉搜索树,使得所有节点的值在[L, R]中 (R>=L) .你可能需要改变树的根节点,所以结果应当返回修剪好的二叉搜索树的新的根节点. 错误的答案: class Solution { public: TreeNode* trimBST(TreeNode* root, int L, int R) { if(root == NULL) return NULL; if(root ->val < L) { root = root…

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…

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 binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree. 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 both v and w…

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 inorder and postorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. 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 Bin…

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. 求二叉树的最大深度问题用到深度优先搜索DFS,递归的完美应用,跟求二叉树的最小深度问题原理相同.代码如下: C++ 解法一: class Solution { public: in…

Given n, generate all structurally unique BST's (binary search trees) that store values 1...n. For example,Given n = 3, your program should return all 5 unique BST's shown below. 1 3 3 2 1 \ / / / \ \ 3 2 1 1 3 2 / / \ \ 2 1 2 3 confused what "{1,#,2…

Given a binary tree, return the inorder traversal of its nodes' values. For example:Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,3,2]. Note: Recursive solution is trivial, could you do it iteratively? confused what "{1,#,2,3}" means? > re…

03-树1. List Leaves (25) Given a tree, you are supposed to list all the leaves in the order of top down, and left to right. Input Specification: Each input file contains one test case. For each case, the first line gives a positive integer N (<=10) wh…

4月份很快就过半了,最近都在看WPF,有点落伍了...本来想写一点读书笔记的,还没想好要怎么写.所以为了能够达到每月一篇博客的目标,今天先说一个LeetCode上的面试题:Unique Binary Search Trees. 题目简单翻译过来就是说: 给定n个不同节点,问:可以构造出多少种异构的二叉搜索树.比方说n=3时有5种: 3          3       3       2       1     \        /       /       /  \       \     …