【二叉查找树】02不同的二叉查找树个数II【Unique Binary Search Trees II】
提到二叉查找树,就得想到二叉查找树的递归定义,
左子树的节点值都小于根节点,右子树的节点值都大于根节点。
++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
给定一个n,问有多少个不同的二叉查找树,使得每个节点的值为 1...n?
例如,
给定n=3,你的程序应该返回所有的这5个不同的二叉排序树的个数。
1 3 3 2 1
\ / / / \ \
3 2 1 1 3 2
/ / \ \
2 1 2 3
++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
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
++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
test.cpp:
|
1
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 |
#include <iostream>
#include <cstdio> #include <stack> #include <vector> #include "BinaryTree.h" using namespace std; /** //对每个节点做root节点做遍历判断,当某个节点为root时候满足条件的二叉树可能有多个
for (int k = start; k <= end; ++k) { vector<TreeNode *> leftSubTree = generate(start, k - 1); vector<TreeNode *> rightSubTree = generate(k + 1, end); for (int i = 0; i < leftSubTree.size(); ++i) { for (int j = 0; j < rightSubTree.size(); ++j) { TreeNode *tmp = new TreeNode(k); tmp->left = leftSubTree[i]; tmp->right = rightSubTree[j]; subTree.push_back(tmp); } } } return subTree; } vector<TreeNode *> generateTrees(int n) vector<vector<int> > levelOrder(TreeNode *root) vector<vector<int> > matrix; vector<TreeNode *> path; int count = 1; if(count == 0) int main() vector<TreeNode *> vRoot; vRoot = generateTrees(3); for (int n = 0; n < vRoot.size(); ++n) for (int i = 0; i < vRoot.size(); ++i) |
|
1
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 |
#ifndef _BINARY_TREE_H_
#define _BINARY_TREE_H_ struct TreeNode TreeNode *CreateBinaryTreeNode(int value); #endif /*_BINARY_TREE_H_*/ |
|
1
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 |
#include <iostream>
#include <cstdio> #include "BinaryTree.h" using namespace std; /** //创建结点 return pNode; //连接结点 //打印节点内容以及左右子结点内容 if(pNode->left != NULL) if(pNode->right != NULL) printf("\n"); //前序遍历递归方法打印结点内容 if(pRoot != NULL) if(pRoot->right != NULL) void DestroyTree(TreeNode *pRoot) delete pRoot; DestroyTree(pLeft); |
【二叉查找树】02不同的二叉查找树个数II【Unique Binary Search Trees II】的更多相关文章
- [Swift]LeetCode95. 不同的二叉搜索树 II | Unique Binary Search Trees II
Given an integer n, generate all structurally unique BST's (binary search trees) that store values 1 ...
- [LeetCode] 95. Unique Binary Search Trees II(给定一个数字n,返回所有二叉搜索树) ☆☆☆
Unique Binary Search Trees II leetcode java [LeetCode]Unique Binary Search Trees II 异构二叉查找树II Unique ...
- leetcode 96. Unique Binary Search Trees 、95. Unique Binary Search Trees II 、241. Different Ways to Add Parentheses
96. Unique Binary Search Trees https://www.cnblogs.com/grandyang/p/4299608.html 3由dp[1]*dp[1].dp[0]* ...
- 【LeetCode】95. Unique Binary Search Trees II 解题报告(Python)
[LeetCode]95. Unique Binary Search Trees II 解题报告(Python) 标签(空格分隔): LeetCode 作者: 负雪明烛 id: fuxuemingzh ...
- 【LeetCode】95. Unique Binary Search Trees II
Unique Binary Search Trees II Given n, generate all structurally unique BST's (binary search trees) ...
- 【leetcode】Unique Binary Search Trees II
Unique Binary Search Trees II Given n, generate all structurally unique BST's (binary search trees) ...
- 41. Unique Binary Search Trees && Unique Binary Search Trees II
Unique Binary Search Trees Given n, how many structurally unique BST's (binary search trees) that st ...
- LeetCode: Unique Binary Search Trees II 解题报告
Unique Binary Search Trees II Given n, generate all structurally unique BST's (binary search trees) ...
- Unique Binary Search Trees,Unique Binary Search Trees II
Unique Binary Search Trees Total Accepted: 69271 Total Submissions: 191174 Difficulty: Medium Given ...
- LeetCode解题报告—— Reverse Linked List II & Restore IP Addresses & Unique Binary Search Trees II
1. Reverse Linked List II Reverse a linked list from position m to n. Do it in-place and in one-pass ...
随机推荐
- Android分享图片失败解决方案
前言:在做图片分享到微博或是用彩信分享的时候,会遇到“无法将图片添加到信息中”,其实这个问题的原因是创建的那个图片默认是,只能被当前应用调用,无法被其他应用调用,即分享的时候,无法读取到图片,并提示I ...
- EasyDSS流媒体解决方案实现的实时数据统计报表、视频文件上传、点播、分享、集成代码等功能
之前的EasyDSS作为rtmp流媒体服务器自从推出就备受用户好评,随着用户的需求的变更产品自身的发展是必须的: 为了更好的用户体验和和功能的完善,我们在EasyDSS的基础上增添了服务器硬件数据报表 ...
- 九度OJ 1354:和为S的连续正数序列 (整除)
时间限制:2 秒 内存限制:32 兆 特殊判题:否 提交:2028 解决:630 题目描述: 小明很喜欢数学,有一天他在做数学作业时,要求计算出9~16的和,他马上就写出了正确答案是100.但是他并不 ...
- Android笔记之自定义的RadioGroup、RadioButton,以及View实例状态的保存与恢复
效果图 activity_main.xml <?xml version="1.0" encoding="utf-8"?> <LinearLay ...
- (3)mac下"-bash: mysql: command not found"解决方案
针对 mysql: command not found 输入命令 $ ln -s /usr/local/mysql/bin/mysql /usr/bin 假如你人品不好,被打脸了,提示你权限不够: l ...
- 用cocos2d-html5做的消除类游戏《英雄爱消除》(4)——游戏结束
游戏结束界面: 在前面几个教程中,这个界面的创作所需要的知识点基本我们都讲过了,这里就说下用户数据的缓存吧,也是先来看下源码 /** * Power by html5中文网(html5china.co ...
- Jooq比较偏的用法
count public Integer count(Integer id) { return dslContext.selectCount().from(Tables.<table_name& ...
- Data Structure Array: Find the minimum distance between two numbers
http://www.geeksforgeeks.org/find-the-minimum-distance-between-two-numbers/ #include <iostream> ...
- css目录
想看css文章目录? 点击我
- cat echo 打印菜单
cat << END ============================= 1.apple 2.pear 3.banana ...