B-tree/B+tree/B*tree

详见:剑指 Offer 题目汇总索引:第6题 Binary Tree Postorder Traversal            Given a binary tree, return the postorder traversal of its nodes' values. For example: Given binary tree {1,#,2,3}, 1 2 / 3 return [3,2,1]. Note: Recursive solution is trivial, could y

简单来说,就是二叉树的前序.中序.后序遍历,包括了递归和非递归的方法 前序遍历(注释中的为递归版本): 1 #include <vector> 2 #include <stack> 3 #include <stddef.h> 4 #include <iostream> 5 6 using namespace std; 7 8 struct TreeNode 9 { 10 int val; 11 TreeNode *left; 12 TreeNode *rig

Creating a Flex Tree with variable height rows can be mysteriously difficult if you are new to Flex item renderers.  This cookbook will give you the step by step instructions to creating a Tree control with variable height renderers. First you will n

计算树的深度 1.minimum depth of binary tree 1 /** 2 * Definition for binary tree 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public:

Binary Tree Preorder Traversal 题目链接 题目要求: Given a binary tree, return the preorder traversal of its nodes' values. For example: Given binary tree {1,#,2,3}, 1 2 / 3 return [1,2,3]. Note: Recursive solution is trivial, could you do it iteratively? 递归解

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,null,null,15,7], 3   / \  9  20    /  \   15   7 return its level order traversal as: [  [3], 

Tree Time Limit: 5000MS   Memory Limit: 131072K Total Submissions: 6131   Accepted: 1682 Description You are given a tree with N nodes. The tree's nodes are numbered 1 through N and its edges are numbered 1 through N ? 1. Each edge is associated with

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. 二叉树的深度,用递归特别方便! 代码如下: public int maxDepth(TreeNode root) { if(root==null){ return 0; } int

平衡树 1.定义 对于每个结点,左右两个子树的高度差的绝对值不超过1,或者叫深度差不超过1 为什么会出现这样一种树呢? 假如我们按照1-n的顺序插入到二叉排序树中,那么二叉排序树就退化成了一个有序链表,效率大大降低. 2.有关概念 所有平衡树基本由以下三个特征组成: 1.自平衡条件 2.旋转操作 3.旋转的触发 平衡树通过设置合理的自平衡条件,使得二叉排序树的查找.插入等操作的性能不至于退化到 O(n)O(n),并且在进行二叉排序树的查找.插入等操作时进行判断,如果满足其中某个旋转的触发条件,则

Given a binary tree, return the postorder traversal of its nodes' values. For example: Given binary tree {1,#,2,3}, 1 2 / 3 return [3,2,1]. Note: Recursive solution is trivial, could you do it iteratively? class Solution { public: vector<int> postor