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. 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 ...

  2. 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 ...

  3. 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 ...

  4. 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 ...

  5. (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 ...

  6. 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 ...

  7. Jooq比较偏的用法

    count public Integer count(Integer id) { return dslContext.selectCount().from(Tables.<table_name& ...

  8. Data Structure Array: Find the minimum distance between two numbers

    http://www.geeksforgeeks.org/find-the-minimum-distance-between-two-numbers/ #include <iostream> ...

  9. css目录

    想看css文章目录? 点击我

  10. cat echo 打印菜单

    cat << END        =============================        1.apple        2.pear        3.banana   ...