<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.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    unordered_map<int , TreeNode*>m;
    
    void rec(vector<int>&nums , int l , int h , int prevmid , bool &f)
    {
        if(l>h) return;
        int mid = (l + h) / 2;
        int present_mid = nums[mid];
        TreeNode* n = new TreeNode(present_mid);
        if(f)
        {
            m[present_mid] = n;
            if(present_mid >= prevmid)
            {
                m[prevmid]->right = n;
            }
            else m[prevmid]->left = n;
        }
        f = true;
        
        rec(nums , l , mid - 1 , present_mid , f);
        rec(nums , mid + 1 , h , present_mid , f);
        
    }
    TreeNode* sortedListToBST(ListNode* head) {
        vector<int>nums;
        while(head)
        {
            nums.push_back(head->val);
            head = head->next;
        }

        int l = 0 , h = nums.size() - 1;
        bool f = false;

        if(h == -1)
        {
            return NULL;
        }
        
        int root = nums[(l + h)/2];
        TreeNode* nn = new TreeNode(root);
        
        m[root] = nn;
        
        rec(nums , l , h , root , f);
        return m[root];
    }
};
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