题目链接

108.将有序数组转换为二叉搜索树

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

小结：不简单的简单题，要想到以数组首尾位置作为参数，递归进行建树。