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

Solution1.a：Recursive

思路：mid處建立Node，mid左右遞歸建立左右樹
Time Complexity: O(N) Space Complexity: O(logN) 遞歸緩存

Solution1.b：Recursive

將end作為出界中止位置，而不是最后一個(gè)元素
與109同步

Solution1.a Code：

class Solution {
    public TreeNode sortedArrayToBST(int[] nums) {
        if (nums.length == 0) {
            return null;
        }
        TreeNode root = helper(nums, 0, nums.length - 1);
        return root;
    }
    
    private TreeNode helper(int[] nums, int start, int end) {
        if(start > end) return null;
        
        int mid = start + (end - start) / 2;
        TreeNode node = new TreeNode(nums[mid]);
        node.left = helper(nums, start, mid - 1);
        node.right = helper(nums, mid + 1, end);
        return node;
    }
}

Solution1.b Code：

class Solution {
    public TreeNode sortedArrayToBST(int[] nums) {
        if(nums == null || nums.length == 0) return null;
        return buildTree(nums, 0, nums.length);
    }
    
    private TreeNode buildTree(int[] nums, int start, int end) {
        if(start == end) return null;

        int mid = start + (end - start) / 2;
        TreeNode root = new TreeNode(nums[mid]);
        
        root.left = buildTree(nums, start, mid);
        root.right = buildTree(nums, mid + 1, end);
        
        return root;
    }
}