<p>Given the <code>head</code> of a singly linked list where elements are sorted in <strong>ascending order</strong>, convert <em>it to a </em><span data-keyword="height-balanced"><strong><em>height-balanced</em></strong></span> <em>binary search tree</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/08/17/linked.jpg" style="width: 500px; height: 388px;" />
<pre>
<strong>Input:</strong> head = [-10,-3,0,5,9]
<strong>Output:</strong> [0,-3,9,-10,null,5]
<strong>Explanation:</strong> One possible answer is [0,-3,9,-10,null,5], which represents the shown height balanced BST.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> head = []
<strong>Output:</strong> []
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in <code>head</code> is in the range <code>[0, 2 * 10<sup>4</sup>]</code>.</li>
	<li><code>-10<sup>5</sup> &lt;= Node.val &lt;= 10<sup>5</sup></code></li>
</ul>

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
# 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 sortedListToBST(self, head: Optional[ListNode]) -> Optional[TreeNode]:
        # this can be done recursively by
        # always taking the middle
        # and then taking the next middles
        # and next middles
        # etc etc
        if head is None:
            return None
        nodes = []
        while head is not None:
            nodes.append(head)
            head = head.next

        q= deque(maxlen=len(nodes))
        q.append((0, len(nodes)-1))
        while len(q) > 0:
            # for each item in the current level,
            # set its left and right
            start,end = q.popleft()
            middle = (start+end)//2
            node = nodes[middle]
            # set left
            if start > middle-1:
                node.left = None
            else:
                node.left = nodes[(start+middle-1)//2]
                q.append((start, middle-1))
            
            # set right
            if middle+1 > end:
                node.right = None
            else:
                node.right = nodes[(middle+1 + end)//2]
                q.append((middle+1, end))
        
        return nodes[(0 + len(nodes)-1)//2]
20200