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;

/**
 * Definition for binary tree with next pointer.
 * struct TreeLinkNode {
 *  int val;
 *  TreeLinkNode *left, *right, *next;
 *  TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
 * };
 */
vector<TreeNode *> generate(int start, int end)
{
    vector<TreeNode *> subTree;
    if (start > end)
    {
        subTree.push_back(NULL);
        return subTree;
    }

    //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)
{
    if (n == 0)
    {
        return generate(1, 0);
    }
    return generate(1, n);
}

vector<vector<int> > levelOrder(TreeNode *root)
{

vector<vector<int> > matrix;
    if(root == NULL)
    {
        return matrix;
    }
    vector<int> temp;
    temp.push_back(root->val);
    matrix.push_back(temp);

vector<TreeNode *> path;
    path.push_back(root);

int count = 1;
    while(!path.empty())
    {
        TreeNode *tn = path.front();
        if(tn->left)
        {
            path.push_back(tn->left);
        }
        if(tn->right)
        {
            path.push_back(tn->right);
        }
        path.erase(path.begin());
        count--;

if(count == 0)
        {
            vector<int> tmp;
            vector<TreeNode *>::iterator it = path.begin();
            for(; it != path.end(); ++it)
            {
                tmp.push_back((*it)->val);
            }
            if(tmp.size() > 0)
            {
                matrix.push_back(tmp);
            }
            count = path.size();
        }
    }
    return matrix;
}

int main()
{

vector<TreeNode *> vRoot;
    vector<vector<int> > ans;

vRoot = generateTrees(3);

for (int n = 0; n < vRoot.size(); ++n)
    {
        ans.clear();
        ans = levelOrder(vRoot[n]);
        cout << "----------------------" << endl;
        for (int i = 0; i < ans.size(); ++i)
        {
            for (int j = 0; j < ans[i].size(); ++j)
            {
                cout << ans[i][j] << " ";
            }
            cout << endl;
        }
    }

for (int i = 0; i < vRoot.size(); ++i)
    {
        DestroyTree(vRoot[i]);
    }
    return 0;
}

 
结果输出:
----------------------
1
2
3
----------------------
1
3
2
----------------------
2
1 3
----------------------
3
1
2
----------------------
3
2
1
 
BinaryTree.h:
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
{
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

TreeNode *CreateBinaryTreeNode(int value);
void ConnectTreeNodes(TreeNode *pParent,
                      TreeNode *pLeft, TreeNode *pRight);
void PrintTreeNode(TreeNode *pNode);
void PrintTree(TreeNode *pRoot);
void DestroyTree(TreeNode *pRoot);

#endif /*_BINARY_TREE_H_*/

BinaryTree.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
 
#include <iostream>
#include <cstdio>
#include "BinaryTree.h"

using namespace std;

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */

//创建结点
TreeNode *CreateBinaryTreeNode(int value)
{
    TreeNode *pNode = new TreeNode(value);

return pNode;
}

//连接结点
void ConnectTreeNodes(TreeNode *pParent, TreeNode *pLeft, TreeNode *pRight)
{
    if(pParent != NULL)
    {
        pParent->left = pLeft;
        pParent->right = pRight;
    }
}

//打印节点内容以及左右子结点内容
void PrintTreeNode(TreeNode *pNode)
{
    if(pNode != NULL)
    {
        printf("value of this node is: %d\n", pNode->val);

if(pNode->left != NULL)
            printf("value of its left child is: %d.\n", pNode->left->val);
        else
            printf("left child is null.\n");

if(pNode->right != NULL)
            printf("value of its right child is: %d.\n", pNode->right->val);
        else
            printf("right child is null.\n");
    }
    else
    {
        printf("this node is null.\n");
    }

printf("\n");
}

//前序遍历递归方法打印结点内容
void PrintTree(TreeNode *pRoot)
{
    PrintTreeNode(pRoot);

if(pRoot != NULL)
    {
        if(pRoot->left != NULL)
            PrintTree(pRoot->left);

if(pRoot->right != NULL)
            PrintTree(pRoot->right);
    }
}

void DestroyTree(TreeNode *pRoot)
{
    if(pRoot != NULL)
    {
        TreeNode *pLeft = pRoot->left;
        TreeNode *pRight = pRoot->right;

delete pRoot;
        pRoot = NULL;

DestroyTree(pLeft);
        DestroyTree(pRight);
    }
}

 
 
 
 

【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

  1. [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 ...

  2. [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 ...

  3. 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]* ...

  4. 【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 ...

  5. 【LeetCode】95. Unique Binary Search Trees II

    Unique Binary Search Trees II Given n, generate all structurally unique BST's (binary search trees) ...

  6. 【leetcode】Unique Binary Search Trees II

    Unique Binary Search Trees II Given n, generate all structurally unique BST's (binary search trees) ...

  7. 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 ...

  8. 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) ...

  9. Unique Binary Search Trees,Unique Binary Search Trees II

    Unique Binary Search Trees Total Accepted: 69271 Total Submissions: 191174 Difficulty: Medium Given  ...

  10. 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

  1. vue实践---vue不依赖外部资源实现简单多语

    vue使用多语,最常见的就是 vue-i18n, 但是如果开发中的多语很少,比如就不到10个多语,这样就没必要引入vue-i18n了, 引入了反正导致代码体积大了,这时候单纯用vue实现多语就是比较好 ...

  2. ULN2003A 使用,有坑

    8脚接24V负极 9脚接24V正极 16接24V继电器,再接到24V正极 1-7无论给5V 正 或 负,10-16都不能达到24V,越靠近输入端的输出端电压越大,最大的才11V,最小的2.5V 最后发 ...

  3. apache虚拟主机配置: 设置二级目录访问跳转

    <VirtualHost *:> DocumentRoot "d:/www/abc" ServerName www.abc.com Alias /course &quo ...

  4. 【Android】开发优化之——调优工具:dump hprof file 查看内存情况,找到内存泄露

    虽说知道一般性的开发android应用须要注意的问题,但是也有水平參差不齐的情况.特别是维护代码,假设内存占用大,内存溢出严重,又怎么解决呢?  --  通过DDMS把heap抓出来分析 1.打开DD ...

  5. CentOS 7.0 systemd

    CentOS 7 已经切换到 systemd,系统指令也有所变化.之前用于启动.重启.停止各种服务的service 作为向后兼容的指令还能使用,但是将来可能会消失.同时,chkconfig 也改成了s ...

  6. sin6_addr打印:string to sockaddr_in6 and sockaddr_in6 to string

    函式原型: #include <arpa/inet.h> const char *inet_ntop(int af, const void *src, char *dst, socklen ...

  7. 【python】将excel转成json

    excel格式如下: 转换后如下 {"BD": 1375.0, "BE": 829.0, "BF": 3.0, "BG" ...

  8. C# 串口调试助手源码

    本方法,禁用跨进程错误(做法不太好,但是对于单片机出身的人来说,好理解,能用就行). 基本功能: 1.点串口号的下拉菜单自动当前检索设备管理器的COM 2.发送模式可选,hex和string两种 3. ...

  9. 关于树莓派Pi2通过UART连接攀藤G5传感器的python

    1.准备工作:树莓派Pi2板子,攀藤G5传感器 关于树莓派40pin口网上很多,我们只了解与攀藤G5连接的问题 (1)攀藤G5pin1(VCC5v)要注意是5V,有很多板子接的是3V,而树莓派的pin ...

  10. python链表的实现

    根据Problem Solving with Algorithms and Data Structures using Python 一书用python实现链表 书籍在线网址http://intera ...