面试题22：有序数组生成不同结构BST

1. 对于一个含有n个数的有序数组1~N，能够产生多少种不同结果的二叉搜素树BST？
2. 如何生成这些不同结构的BST？
3. 有序数组如何生成平衡二叉搜索树？

 class Solution {
 public:
     int numTrees(int n) {
         int* dp = new int[n+];
         dp[] = ;
         dp[] = ;
         dp[] = ;
         for(int i=;i<=n;i++){
             dp[i] = ;
             for(int j=;j<=i;j++){
                 dp[i]+=dp[j-]*dp[i-j];
             }
         }
         return dp[n];
     }

     int numTrees2(int n){
         if(n==) return ;
         if(n < ) return n;
         int sum = ;
         for(int i=;i<=n;i++){
             sum += numTrees2(i-)*numTrees2(n-i);
         }
         return sum;
     }
 };

 /**
  * 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 *> generateTrees(int left, int right) {
         vector<TreeNode*> res;
         if (left > right) {
             res.push_back(nullptr);
             return res;
         }

         if(left == right){
             res.push_back(new TreeNode(left));
             return res;
         }

         for (int i = left; i <= right; i++) {
             //TreeNode* root = new TreeNode(i);
             vector<TreeNode*> leftRes = generateTrees(left, i-);
             vector<TreeNode*> rightRes = generateTrees(i + , right);
             for (size_t j = ; j < leftRes.size(); j++) {
                 for (size_t k = ; k < rightRes.size(); k++) {
                     TreeNode* root = new TreeNode(i);
                     root->left = leftRes[j];
                     root->right = rightRes[k];
                     res.push_back(root);
                 }
             }
         }
         return res;
     }

     vector<TreeNode*> generateTrees(int n){
         vector<TreeNode*> res;
         if(n == ) {
             res.push_back(nullptr);
             return res;
         }
         return generateTrees(,n);
     }

 };