Convert Sorted Array to Binary Search Tree
<p>Given an integer array <code>nums</code> where the 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/0100-0199/0108.Convert%20Sorted%20Array%20to%20Binary%20Search%20Tree/images/btree1.jpg" style="width: 302px; height: 222px;" />
<pre>
<strong>Input:</strong> nums = [-10,-3,0,5,9]
<strong>Output:</strong> [0,-3,9,-10,null,5]
<strong>Explanation:</strong> [0,-10,5,null,-3,null,9] is also accepted:
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0100-0199/0108.Convert%20Sorted%20Array%20to%20Binary%20Search%20Tree/images/btree2.jpg" style="width: 302px; height: 222px;" />
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0100-0199/0108.Convert%20Sorted%20Array%20to%20Binary%20Search%20Tree/images/btree.jpg" style="width: 342px; height: 142px;" />
<pre>
<strong>Input:</strong> nums = [1,3]
<strong>Output:</strong> [3,1]
<strong>Explanation:</strong> [1,null,3] and [3,1] are both height-balanced BSTs.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>4</sup></code></li>
	<li><code>-10<sup>4</sup> &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
	<li><code>nums</code> is sorted in a <strong>strictly increasing</strong> order.</li>
</ul>
Tree; Binary Search Tree; Array; Divide and Conquer; Binary Tree
/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func sortedArrayToBST(nums []int) *TreeNode {
	var dfs func(int, int) *TreeNode
	dfs = func(l, r int) *TreeNode {
		if l > r {
			return nil
		}
		mid := (l + r) >> 1
		return &TreeNode{nums[mid], dfs(l, mid-1), dfs(mid+1, r)}
	}
	return dfs(0, len(nums)-1)
}
