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

一刷
題解：recursion, 思路很簡(jiǎn)單

/**
 * 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) return null;
        return sortedArrayToBST(nums, 0, nums.length-1);
    }
    
    private TreeNode sortedArrayToBST(int[] nums, int start, int end){
        if(start<0 || end>nums.length-1) return null;
        if(start>end) return null;
        int mIndex = start + (end-start)/2;
        TreeNode root = new TreeNode(nums[mIndex]);
        root.left = sortedArrayToBST(nums, start, mIndex-1);
        root.right = sortedArrayToBST(nums, mIndex+1, end);
        return root;
    }
}

二刷
recursion

/**
 * 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) {
        return sortedArrayToBST(nums, 0, nums.length-1);
    }
    
    public TreeNode sortedArrayToBST(int[] nums, int lo, int hi){
        if(hi<lo) return null;
        int mid = lo + (hi - lo)/2;
        TreeNode root = new TreeNode(nums[mid]);
        root.left = sortedArrayToBST(nums, lo, mid-1);
        root.right = sortedArrayToBST(nums, mid+1, hi);
        return root;
    }
}