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# 108.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]`
/**
* 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

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;
}
}