Leetcode 95 不同的二叉搜索树II

题目描述：

 

 题解：首先要了解卡特兰数的递归公式。对于序列[1,2,....n]构成的二叉搜索树，我们按顺序选择i作为根节点，[1,..i-1]构成左子树，[i+1,....n]作为右子树。递归的重复上述操作即可

（之前写树的题返回值都是单个值，这里递归返回值为节点序列，因为有很多种情况）

AC代码：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<TreeNode*> Build(int low,int hight)
    {
        if(low > hight) return {};
        vector<TreeNode* > tmp;
        // if(low == hight)
        // {
        //     TreeNode* now = new TreeNode(low);
        //     tmp.push_back(now);
        //     return tmp;
        // }
        for(int i=low;i<=hight;i++)
        {
            vector<TreeNode*> l = Build(low,i-1);
            vector<TreeNode*> r = Build(i+1,hight);
            int Len_l = l.size();
            int Len_r = r.size();
            if(Len_l == 0)
            {
                l.push_back(NULL);
                Len_l = l.size();
            } 
            if(Len_r == 0) 
            {
                r.push_back(NULL);
                Len_r = r.size();
            }
            
            for(int j = 0;j < Len_l;j++)
            {
                for(int k = 0;k < Len_r;k++)
                {
                    TreeNode* now = new TreeNode(i);
                    now->left = l[j];
                    now->right = r[k];
                    tmp.push_back(now);
                }
            }
        }
        return tmp;
    }

    vector<TreeNode*> generateTrees(int n) {
        vector<TreeNode* > ans;
        ans = Build(1,n);        
        return ans;
    }
};