108. Convert Sorted Array to Binary Search Tree 将有序数组转换为二叉搜索树
作者: 负雪明烛 id: fuxuemingzhu 个人博客: https://fuxuemingzhu.cn/
题目地址:https://leetcode.com/problems/convert-sorted-array-to-binary-search-tree/#/description
Given an array where elements are sorted in ascending order, convert it to a height balanced BST.
For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.
Example:
Given the sorted array: [-10,-3,0,5,9],
One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:
0 / \ -3 9 / / -10 5把一个已经排序了的数组,变成一个高度平衡的BST。答案不唯一。
Java解法
Section titled “Java解法”因为BST的中序遍历是有序的,所以有序数组的中间的数字是根节点,序列中间节点左边是根节点的左子树,右边是根节点的右子树,以此类推。
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */public class Solution { public TreeNode sortedArrayToBST(int[] nums) { return helper(nums, 0, nums.length - 1); }
public TreeNode helper(int[] nums, int start, int end){ if(start > end){ return null; } int mid = (start + end) / 2; TreeNode node = new TreeNode(nums[mid]); node.left = helper(nums, start, mid - 1); node.right = helper(nums, mid + 1, end); return node; }}Python解法
Section titled “Python解法”二刷,python
用python2的时候,最后有个特别大的测试用例,导致内存错误。。
改成Python3,并把除法改成了地板除竟然过了。。神奇。
# Definition for a binary tree node.# class TreeNode(object):# def __init__(self, x):# self.val = x# self.left = None# self.right = None
class Solution(object): def sortedArrayToBST(self, nums): """ :type nums: List[int] :rtype: TreeNode """ if not nums: return None _len = len(nums) mid = _len // 2 root = TreeNode(nums[mid]) root.left = self.sortedArrayToBST(nums[:mid]) root.right = self.sortedArrayToBST(nums[mid+1:]) return root2017 年 4 月 24 日 2018 年 6 月 23 日 2018 年 11 月 16 日 —— 又到周五了!

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