LeetCode 108. Convert Sorted Array to Binary Search Tree-优快云博客

本文介绍如何将一个已排序的数组转换为一个平衡的二叉搜索树，详细阐述了实现过程和时间空间复杂度。

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原题链接在这里：https://leetcode.com/problems/convert-sorted-array-to-binary-search-tree/

题目：

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

题解：

本题与Convert Sorted List to Binary Search Tree相似. 为了方便记左右index来做递归，建一个有多个argument的函数。这也是array递归的通用方法。然后中点取值返回TreeNode, 左侧用作生成left subtree, 右侧用作生成right subtree.

Time Complexity: O(n), 以为每个点只访问了一次. Space: O(logn), 递归用了O(logn)层stack.

AC Java:

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 class Solution {
11     public TreeNode sortedArrayToBST(int[] nums) {
12         if(nums == null || nums.length == 0){
13             return null;
14         }
15         
16         return transform(nums, 0, nums.length-1);
17     }
18     
19     private TreeNode transform(int [] nums, int left, int right){
20         if(left > right){
21             return null;
22         }
23         
24         int mid = left + (right-left)/2;
25         TreeNode root = new TreeNode(nums[mid]);
26         root.left = transform(nums, left, mid-1);
27         root.right = transform(nums, mid+1, right);
28         
29         return root;
30     }
31 }