<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 {
 public:
  TreeNode* sortedListToBST(ListNode* head) {
    if (head == nullptr)
      return nullptr;
    if (!head->next)
      return new TreeNode(head->val);

    ListNode* mid = findMid(head);
    TreeNode* root = new TreeNode(mid->val);
    root->left = sortedListToBST(head);
    root->right = sortedListToBST(mid->next);
    return root;
  }

 private:
  ListNode* findMid(ListNode* head) {
    ListNode* prev = nullptr;
    ListNode* slow = head;
    ListNode* fast = head;

    while (fast != nullptr && fast->next != nullptr) {
      prev = slow;
      slow = slow->next;
      fast = fast->next->next;
    }
    prev->next = nullptr;

    return slow;
  }
};
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