LC.95.不同的二叉搜索树 II(递归)

LC.95.不同的二叉搜索树 II(递归)

思路：递归实现，考虑枚举根结点，递归左右子树，然后从左子树的答案和右子树的答案选一个合并丢进 a n s ans ans里即可。

/**
 * 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*> dfs(int l,int r){
        if(l>r) return  {nullptr};
        vector<TreeNode*> ans;
        for(int i=l;i<=r;i++){
            vector<TreeNode*> left_tree=dfs(l,i-1);
            vector<TreeNode*> right_tree=dfs(i+1,r);
            for(auto j:left_tree)
                for(auto k:right_tree){
                    TreeNode * Cur=new TreeNode(i);
                    Cur->left=j;
                    Cur->right=k;
                    ans.push_back(Cur);
                }
        }
        return ans;
    }
    vector<TreeNode*> generateTrees(int n) {
         if(!n) return {};
         return dfs(1,n);
    }
};