<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]]:
        '''
        def dfs(start, end):
            if (start, end) in cache: return cache[(start, end)]
            
            cache[(start, end)] = []

            # edge case
            if start > end:
                cache[(start, end)].append(None)
                return cache[(start, end)]
            if start == end:
                cache[(start, end)].append(TreeNode(start))
                return cache[(start, end)]

            # 
            for i in range(start, end+1):
                left = dfs(start, i-1)
                right = dfs(i+1, end)
            
                for cur_left in left:
                    for cur_right in right:
                        cache[(start, end)].append(TreeNode(i, cur_left, cur_right))
            return cache[(start, end)]

        cache = {}

        return dfs(1, n)
        '''

        def dfs(l, r):
            if (l, r) in cache: return cache[(l, r)]
            if l == r: return [TreeNode(val=l)]
            if l > r: return [None]
            # if l > r: return None
            ans = []
            for m in range(l, r+1):
                for left in dfs(l, m-1):
                    for right in dfs(m+1, r):
                        ans.append(TreeNode(val=m, left=left, right=right))
            cache[(l, r)] = ans
            return cache[(l, r)] # all trees
        
        cache = {}
        return dfs(1, n)
17500