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：

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<TreeNode *>builtree(int left,int right)
    {
        vector<TreeNode *>data;
        if(left>right)
        {
            data.push_back(nullptr);
            return data;
        }
        
        vector<TreeNode *>leftr,rightr;
        for(int i=left;i<=right;i++)
        {
         
            leftr = builtree(left,i-1);
            rightr = builtree(i+1,right);
            for(int t1=0;t1<leftr.size();t1++)
                for(int t2=0;t2<rightr.size();t2++)
                {
                    TreeNode *nod = new TreeNode(i);
                    nod->left = leftr[t1];
                    nod->right = rightr[t2];
                    data.push_back(nod);
                }
        }
        return data;
        
    }
    vector<TreeNode *> generateTrees(int n) {
        vector<TreeNode *> data(1,nullptr);
        data = builtree(1,n);
        return data;
    }
};