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: http://fuxuemingzhu.cn/
@TOC
题目地址:https://leetcode.com/problems/convert-sorted-array-to-binary-search-tree/#/description
题目描述
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
题目大意
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。
解题方法
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
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 root
日期
2017 年 4 月 24 日 2018 年 6 月 23 日 2018 年 11 月 16 日 —— 又到周五了!