Convert Sorted List to Binary Search Tree
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0100-0199/0109.Convert%20Sorted%20List%20to%20Binary%20Search%20Tree/images/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>
Tree; Binary Search Tree; Linked List; Divide and Conquer; Binary Tree
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
struct TreeNode* dfs(int* nums, int i, int j) {
    if (i > j) {
        return NULL;
    }
    int mid = (i + j) >> 1;
    struct TreeNode* left = dfs(nums, i, mid - 1);
    struct TreeNode* right = dfs(nums, mid + 1, j);
    struct TreeNode* root = (struct TreeNode*) malloc(sizeof(struct TreeNode));
    root->val = nums[mid];
    root->left = left;
    root->right = right;
    return root;
}

struct TreeNode* sortedListToBST(struct ListNode* head) {
    int size = 0;
    struct ListNode* temp = head;
    while (temp) {
        size++;
        temp = temp->next;
    }

    int* nums = (int*) malloc(size * sizeof(int));
    temp = head;
    for (int i = 0; i < size; i++) {
        nums[i] = temp->val;
        temp = temp->next;
    }

    struct TreeNode* root = dfs(nums, 0, size - 1);
    free(nums);
    return root;
}
