问题背景
给你一个整数数组 n u m s nums nums,其中元素已经按 升序 排列,请你将其转换为一棵 平衡 二叉搜索树。
数据约束
- 1 ≤ n u m s . l e n g t h ≤ 1 0 4 1 \le nums.length \le 10 ^ 4 1≤nums.length≤104
- − 1 0 4 ≤ n u m s [ i ] ≤ 1 0 4 -10 ^ 4 \le nums[i] \le 10 ^ 4 −104≤nums[i]≤104
- n u m s nums nums 按 严格递增 顺序排列
解题过程
数组已经升序排列,实际上平衡二叉树每层的节点都是前一层递归中数组的中间节点,依次构造好即可。
具体实现
/*** 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 dfs(nums, 0, nums.length);}private TreeNode dfs(int[] nums, int left, int right) {if(left == right) {return null;}int mid = left + ((right - left) >>> 1);return new TreeNode(nums[mid], dfs(nums, left, mid), dfs(nums, mid + 1, right));}
}