Leetcode - Convert Sorted Array to Binary Search Tree

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My code:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode sortedArrayToBST(int[] nums) {
        if (nums == null || nums.length == 0)
            return null;
        
        return getBST(0, nums.length - 1, nums);
    }
    
    private TreeNode getBST(int begin, int end, int[] nums) {
        if (begin > end)
            return null;
        else if (begin == end)
            return new TreeNode(nums[begin]);
        
        int mid = (begin + end) / 2;
        TreeNode root = new TreeNode(nums[mid]);
        root.left = getBST(begin, mid - 1, nums);
        root.right = getBST(mid + 1, end, nums);
        return root;
    }
}

My test result:

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这道题目和上面的一个想法.具体就不细说了。

**
总结： sorted array -> balanced bst
**

Anyway, Good luck, Richardo!

My code:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode sortedArrayToBST(int[] nums) {
        if (nums == null || nums.length == 0) {
            return null;
        }
        
        return helper(nums, 0, nums.length - 1);
    }
    
    private TreeNode helper(int[] nums, int begin, int end) {
        if (begin > end) {
            return null;
        }
        else if (begin == end) {
            return new TreeNode(nums[begin]);
        }
        int middle = begin + (end - begin) / 2;
        TreeNode root = new TreeNode(nums[middle]);
        root.left = helper(nums, begin, middle - 1);
        root.right = helper(nums, middle + 1, end);
        return root;
    }
}

简单题，没什么好说的。
Anyway, Good luck, Richardo! -- 08/16/2016