<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) {
        // bc
        if(head==NULL) return NULL;
        if(head->next==NULL) return new TreeNode(head->val);

        ListNode* slow=head;
        ListNode* fast= head;
        ListNode* prev=NULL;
        while(fast!=NULL && fast->next!=NULL){
            prev=slow;
            fast=fast->next->next;
            slow=slow->next;
        }
        prev->next=NULL;
        TreeNode* root= new TreeNode(slow->val);
        root->left= sortedListToBST(head);
        root->right= sortedListToBST(slow->next);
        return root;
        
    }
};
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