<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 buildbst(self,stack,rootidx):
        lstack=stack[:rootidx]
        rstack=stack[rootidx+1:]
        lret=[]
        rret=[]
        ret=[]
        if not lstack:
            lret=[None]
        if not rstack:
            rret=[None]
        if len(lstack)==1:
            root=TreeNode()
            root.val=lstack[0]
            root.left=None
            root.right=None
            lret.append(root)
        else:
            for index in range(len(lstack)):
                lret+=self.buildbst(lstack,index)
        if len(rstack)==1:
            root=TreeNode()
            root.val=rstack[0]
            root.left=None
            root.right=None
            rret.append(root)
        else:
            for index in range(len(rstack)):
                rret+=self.buildbst(rstack,index)
        for litm in lret:
            for ritm in rret:
                root=TreeNode()
                root.val=stack[rootidx]
                root.left=litm
                root.right=ritm
                ret.append(root)
        return ret
    def generateTrees(self, n: int) -> List[Optional[TreeNode]]:
        stack=list(range(1,n+1))
        trees=[]
        for idx in range(0,n):
            trees+=self.buildbst(stack,idx)
        return trees
18000