# 108. Convert Sorted Array to Binary Search Tree

**EASY**

Given an integer array `nums` where the elements are sorted in **ascending order**, convert *it to a **height-balanced** binary search tree*.

A **height-balanced** binary tree is a binary tree in which the depth of the two subtrees of every node never differs by more than one.

 

**Example 1:**

![](https://assets.leetcode.com/uploads/2021/02/18/btree1.jpg)

<pre><code>Input: 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:
</code></pre>

**Example 2:**

![](https://assets.leetcode.com/uploads/2021/02/18/btree.jpg)

<pre><code>Input: nums = [1,3]
<strong>Output:
</strong> [3,1]
<strong>Explanation:
</strong> [1,null,3] and [3,1] are both height-balanced BSTs.
</code></pre>

&#x20;

**Constraints:**

* `1 <= nums.length <= 104`
* `-104 <= nums[i] <= 104`
* `nums` is sorted in a **strictly increasing** order.

```python
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def sortedArrayToBST(self, nums: List[int]) -> Optional[TreeNode]:
        length = len(nums)
        if length == 0:
            return None
        index = length // 2
        root = TreeNode(nums[index])
        root.left = self.sortedArrayToBST(nums[:index])
        root.right = self.sortedArrayToBST(nums[index+1:])
        return root
```