将已排好序的数组转换成高度平衡的二叉搜索树(BST)

题目要求：给定一个数组，元素已经按照升序排列，将其转换成高度平衡的二叉搜索树(BST)

/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public TreeNode sortedArrayToBST(int[] num) {
int len = num.length;

if (num.length == 0) { //异常判断
return null;
}

return recurseOpera(num, 0, len - 1);
}
//因为数组已经排好序
//所以每次只需要取出数组的中间位的数据来生成一个节点，然后递归即可
public TreeNode recurseOpera(int[] num, int low, int high) {
if (low > high) { //递归的出口
return null;
}
int mid = (low + high) / 2;
TreeNode tnode = new TreeNode(num[mid]);
tnode.left = recurseOpera(num, low, mid - 1);
tnode.right = recurseOpera(num, mid + 1, high);

return tnode;
}