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==) return ;
        if(n==) return ;
        int wnum = ;
        for(int i=;i<=n;i++)
        {
            wnum += numTrees(i-) * 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-);
             rightr = builtree(i+,right);
             for(int t1=;t1<leftr.size();t1++)
                 for(int t2=;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(,nullptr);
         data = builtree(,n);
         return data;
     }
 };