95. 不同的二叉搜索树 II(LeetCode)(递归)(重点)

95. 不同的二叉搜索树 II
二叉搜索树中序遍历是有序的

枚举中间点k,左子树必然是l,k-1,右子树必然是k+1,r,然后递归求解左右子树各自多少种情况

/**
 * 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<TreeNode*> generateTrees(int n) {
        return dfs(1, n);
    }
    vector<TreeNode*>dfs(int l, int r){
        if(l > r) return {NULL};
        vector<TreeNode*> res;

        for(int i = l; i <= r; ++i){    
            //返回树的形状结构,lef接受所有可能的结构故而是一个集合
            auto left = dfs(l, i - 1), right = dfs(i + 1, r);
            //从左边任取一棵子树,从右边任取一棵子树,都可以拼成当前子树
            for(auto l:left){
                for(auto r:right){
                    auto root = new TreeNode(i);
                    root->left = l, root->right = r;
                    res.push_back(root);
                }
            }
        }
        return res;
    }
};