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# 108.Convert-Sorted-Array-to-Binary-Search-Tree

## 108. Convert Sorted Array to Binary Search Tree

## 题目地址

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

## 题目描述

```
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
```

## 代码

### Approach #1 Preorder Traversal: Always Choose Left Middle Node as a Root

`int p = (left + right) / 2;`

* `root = p`
* `root.left = [left, m-1]`
* `root.right = [m+1, right]`

```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) {
    return helper(nums, 0, nums.length - 1);
  }

  public TreeNode helper(int[] nums, int left, int right) {
    if (left > right) return null;

    // always choose left middle node as a root
    int p = (left + right) / 2;

    // inorder traversal: left -> node -> right
    TreeNode root = new TreeNode(nums[p]);
    root.left = helper(left, p - 1);
    root.right = helper(p + 1, right);
    return root;
  }

}
```

### Approach #2 Preorder Traversel: Always Choose Right Middle Node as a Root

```java
class Solution {
  public TreeNode sortedArrayToBST(int[] nums) {
    this.nums = nums;
    return helper(0, nums.length - 1);
  }

  public TreeNode helper(int left, int right) {
    if (left > right) return null;

    // always choose right middle node as a root
    int p = (left + right) / 2;
    if ((left + right) % 2 == 1) ++p;

    // inorder traversal: left -> node -> right
    TreeNode root = new TreeNode(nums[p]);
    root.left = helper(left, p - 1);
    root.right = helper(p + 1, right);
    return root;
  }
}
```