<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 allPossibleBSt(self,start,end,memo):
        res=[]
        if start>end:
            res.append(None)
            return res
        if (start,end) in memo:
            return memo[(start,end)]
        for i in range(start,end+1):
            leftSubTrees=self.allPossibleBSt(start,i-1,memo)
            rightSubTree=self.allPossibleBSt(i+1,end,memo)
            for left in leftSubTrees:
                for right in rightSubTree:
                    root=TreeNode(i,left,right)
                    res.append(root)
        memo[(start,end)]=res
        return res

    def generateTrees(self, n: int) -> List[Optional[TreeNode]]:
        memo={}
        return self.allPossibleBSt(1,n,memo)
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