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# Convert Sorted Array to Binary Search Tree

Given an array where elements are sorted in ascending order, convert it to a height balanced BST.

For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of\_every\_node never differ by more than 1.

**Example:**

```
Given the sorted array: [-10,-3,0,5,9],

One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:

      0
     / \
   -3   9
   /   /
 -10  5
```

## Analysis

因为数组是排序过的，因此只需要利用递归，将中点设为根节点，左右子树则分别对应被中点分隔的左右两部分有序数组，再递归生成子树即可。

## Solution

```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 || nums.length == 0) {
            return null;
        }
        return buildBST(nums, 0, nums.length - 1);
    }
    TreeNode buildBST(int[] nums, int left, int right) {
        if (left > right) {
            return null;
        }
        int mid = (left + right) / 2;
        TreeNode node = new TreeNode(nums[mid]);
        node.left = buildBST(nums, left, mid - 1);
        node.right = buildBST(nums, mid + 1, right);
        return node;
    }
}
```


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