<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.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def generateTrees(self, n: int) -> List[Optional[TreeNode]]:
        nodes = [i for i in range(1, n+1)]
        self.res = []
        self._helper(nodes,True)
        return self.res


    def _helper(self, node_list, archive):
        if node_list == []:
            return [None]
        tmp_res = []
        for idx, node in enumerate(node_list):
            for left_option in self._helper(node_list[0:idx], False):
                for right_option in self._helper(node_list[idx+1:], False):
                    tmp_root = TreeNode(val = node)
                    tmp_root.left = left_option
                    tmp_root.right = right_option
                    tmp_res.append(tmp_root)
                    if archive:
                        self.res.append(tmp_root)
        return tmp_res


        
        
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