<p>Given an integer <code>n</code>, return <em>the number of 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>.</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> 5
</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;= 19</code></li>
</ul>

class Solution:
    def __init__(self):
        self.table = {0: 1, 1: 1}

    def numTrees(self, n: int) -> int:
        if n in self.table:
            return self.table[n]

        total_comb = 0
        for i in range(1, n + 1): # iterate n, 代表依序以1, 2, ... n為root建bst計算合法數量
            left_subtree_len = i - 1
            right_subtree_len = n - i
            left_comb = self.numTrees(left_subtree_len)
            right_comb = self.numTrees(right_subtree_len)
            total_comb += left_comb * right_comb
        self.table[n] = total_comb

        return total_comb
16400