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+1];
        dp[0] = 1;
        dp[1] = 1;
        dp[2] = 2;
        for(int i=3;i<=n;i++){
            dp[i] = 0;
            for(int j=1;j<=i;j++){
                dp[i]+=dp[j-1]*dp[i-j];
            }
        }
        return dp[n];
    }

    int numTrees2(int n){
        if(n==0) return 1;
        if(n < 3) return n;
        int sum = 0;
        for(int i=1;i<=n;i++){
            sum += numTrees2(i-1)*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-1);
            vector<TreeNode*> rightRes = generateTrees(i + 1, right);
            for (size_t j = 0; j < leftRes.size(); j++) {
                for (size_t k = 0; 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 == 0) {
            res.push_back(nullptr);
            return res;
        }
        return generateTrees(1,n);
    }

};