<p>Given an integer <code>n</code>, return <em>all the structurally unique <strong>BST&#39;</strong>s (binary search trees), which has exactly </em><code>n</code><em> nodes of unique values from</em> <code>1</code> <em>to</em> <code>n</code>. Return the answer in <strong>any order</strong>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/01/18/uniquebstn3.jpg" style="width: 600px; height: 148px;" />
<pre>
<strong>Input:</strong> n = 3
<strong>Output:</strong> [[1,null,2,null,3],[1,null,3,2],[2,1,3],[3,1,null,null,2],[3,2,null,1]]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 1
<strong>Output:</strong> [[1]]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 8</code></li>
</ul>

/**
 * 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*> all(int start,int end){
        if (start >  end) return {0};
        if (start == end) return {new TreeNode(start)};

        vector<TreeNode*> ans;

        for(int i=start;i<=end;i++){
            vector<TreeNode*> left = all(start,i-1);
            vector<TreeNode*> right = all(i+1,end);

            for(int j=0;j<left.size();j++){
                for(int k=0;k<right.size();k++){
                    TreeNode*root = new TreeNode(i);
                    root->left = left[j];
                    root->right = right[k];
                    ans.push_back(root);
                }
            }
        }
        return ans;
    }
    vector<TreeNode*> generateTrees(int n) {
        if (n == 0) return {};
        return all(1,n);
    }
};
19800