# 0108. Convert Sorted Array to Binary Search Tree

<https://leetcode.com/problems/convert-sorted-array-to-binary-search-tree>

## Description

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)

```
**Input:** nums = [-10,-3,0,5,9]
**Output:** [0,-3,9,-10,null,5]
**Explanation:** [0,-10,5,null,-3,null,9] is also accepted:
![](https://assets.leetcode.com/uploads/2021/02/18/btree2.jpg)
```

**Example 2:**

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

```
**Input:** nums = [1,3]
**Output:** [3,1]
**Explanation:** [1,3] and [3,1] are both a height-balanced BSTs.
```

**Constraints:**

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

## ac

```java
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode sortedArrayToBST(int[] nums) {
        if (nums == null) return null;
        return helper(nums, 0, nums.length - 1);
    }
    private TreeNode helper(int[] nums, int s, int e) {
        //exit, null
        if (nums == null || nums.length == 0 || s < 0 || e >= nums.length || s > e)
            return null;
        if (s == e) return new TreeNode(nums[s]);

        // divide and conquer
        int mid = s + (e - s) / 2;
        TreeNode root = new TreeNode(nums[mid]);
        root.left = helper(nums, s, mid - 1);
        root.right = helper(nums, mid + 1, e);

        return root;
    }
}
```


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