Leetcode: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 store values 1...n?

For example,
Given n = 3, there are a total of 5 unique BST's.

   1         3     3      2      1
    \       /     /      / \      \
     3     2     1      1   3      2
    /     /       \                 \
   2     1         2                 3

Unique Binary Search TreesII

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

分析：这个题目是要求BST有多少种排列方式，第一题是只求数量，第二题要求把结构也存储下来。本质都是深搜，第一题显然更简单

因为只要知道当前有多少个数，就可以知道有多少种排列方式了，甚至可以简单的建个存储表，更快一些。而第二题就需要将当前情况

下所有点的排列搞到，返回上一级，上一级再建立更高一层的bst

第一题代码：

class Solution {
public:
    int numTrees(int n) {
        if(n==0) return 1;
        if(n==1) return 1;
        int wnum = 0;
        for(int i=1;i<=n;i++)
        {
            wnum += numTrees(i-1) * numTrees(n-i);
        }
        
        return wnum;
    }
};

第二题代码：

 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:
12     vector<TreeNode *>builtree(int left,int right)
13     {
14         vector<TreeNode *>data;
15         if(left>right)
16         {
17             data.push_back(nullptr);
18             return data;
19         }
20         
21         vector<TreeNode *>leftr,rightr;
22         for(int i=left;i<=right;i++)
23         {
24          
25             leftr = builtree(left,i-1);
26             rightr = builtree(i+1,right);
27             for(int t1=0;t1<leftr.size();t1++)
28                 for(int t2=0;t2<rightr.size();t2++)
29                 {
30                     TreeNode *nod = new TreeNode(i);
31                     nod->left = leftr[t1];
32                     nod->right = rightr[t2];
33                     data.push_back(nod);
34                 }
35         }
36         return data;
37         
38     }
39     vector<TreeNode *> generateTrees(int n) {
40         vector<TreeNode *> data(1,nullptr);
41         data = builtree(1,n);
42         return data;
43     }
44 };

来源：https://www.cnblogs.com/soyscut/p/3783558.html

标签

bst