leetcode95——不同的二叉搜索树

题目链接：https://leetcode-cn.com/problems/unique-binary-search-trees-ii/

思路：首先从根节点进行分析，可能值为i，（1<=i<=n)。根据二叉搜索树的性质，左子树的值可能为1~i-1，右子树的值可能为i+1~n。然后依次递归下去，返回到当前层，则将左右子树的根结点串到当前结点就可以了。

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector generateBinaryTrees(int index1,int index2)
    {
        if(index1>index2)return {nullptr};
        vectorcur;
        for(int i=index1;i<=index2;i++)
        { 
            vectorleft;
            vectorright;
            left=generateBinaryTrees(index1,i-1);
            right=generateBinaryTrees(i+1,index2);
            TreeNode* temp;
            for(auto& l:left)
            {
                for(auto& r:right)
                {
                    TreeNode* temp=new TreeNode(i);
                    temp->left=l;
                    temp->right=r;
                    cur.push_back(temp);
                }
            }
        }
        return cur;
    }
    vector generateTrees(int n) {
        vectorresult;
        if(!n)return {};
        return generateBinaryTrees(1,n);
    }
};