<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>

class Solution:

  def sortedListToBST(self, head: ListNode) -> TreeNode:

    def findMid(head: ListNode) -> ListNode:

      prev = None

      slow = head

      fast = head

      while fast and fast.next:

        prev = slow

        slow = slow.next

        fast = fast.next.next

      prev.next = None

      return slow

    if not head:

      return None

    if not head.next:

      return TreeNode(head.val)

    mid = findMid(head)

    root = TreeNode(mid.val)

    root.left = self.sortedListToBST(head)

    root.right = self.sortedListToBST(mid.next)

    return root
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