<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) return nullptr;
        
        ListNode* dummy = new ListNode(-1);
        dummy->next = head;
        
        ListNode *slow = dummy, *fast = dummy;
        ListNode *preslow;
        
        while(fast && fast->next){
            preslow = slow;
            slow = slow->next;
            fast = fast->next->next;
        }
        
        cout << preslow->val << ", " << slow->val << endl;
        
        
        TreeNode* root = new TreeNode(slow->val);
        
      
        preslow->next = nullptr;
        
        root->left = (slow == head) ? nullptr : sortedListToBST(head);
        
        root->right = sortedListToBST(slow->next);
        
        return root;
    }
};

31900