力扣hot100 - 108、将有序数组转换为二叉搜索树

题目：思路：将数组从最中间分为中间节点，左边数组和右边数组，中间作为根节点，左边作为左子树，右边作为右子树，左子树右子树同样取中间节点作为根节点。

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