<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]:
        if not head: return None
        if not head.next: return TreeNode(head.val)
        mid = self.find_mid(head)
        node = TreeNode(mid.val)
        mid.left = self.sortedListToBST(head)
        mid.right = self.sortedListToBST(mid.next)
        return mid


    def find_mid(self, node):
        prev,curr,nxt = None, node,node
        while nxt and nxt.next:
            prev,curr,nxt = curr, curr.next, nxt.next.next
        prev.next = None
        return curr
        
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