LeetCode解题报告 108. Convert Sorted Array to Binary Search Tree [medium]
2016-12-20 00:44
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题目描述
Given an array where elements are sorted in ascending order, convert it to a height balanced BST.解题思路
题目要求对一个已经排好序的数组来构造一颗平衡二叉树。平衡二叉树指的是对于树上任意一个节点,其左右子树的高度差不超过1。
若两颗子树节点数量相同,且使用同样的方法来进行构造。那么它们的形态应该是相同的,层数也相同。举个例子:
1 / \ X Y / \ / \ X X Y Y
若两颗子树节点数量相差1,则构造出的子树层数至多相差1。举个例子:
1 / \ X Y / \ / X X Z
只需要保证每一次递归处理构造的左右子树节点数量最多不超过1,就可以保证生成的二叉树满足高度平衡的性质。因此,每一次以中位数进行分割,将数组分割为左右两个部分,对左右两个子树执行相同的操作,不断从中点划分,来递归构造二叉树。
复杂度分析
时间复杂度为O(n),空间复杂度为O(n)。代码如下:
class Solution { public: TreeNode* sortedArrayToBST(vector<int>& nums) { if (nums.size()==0) { return NULL; } int medium=nums.size()/2; vector<int>leftlist(nums.begin(),nums.begin()+medium); vector<int>righlist(nums.begin()+medium+1,nums.end()); TreeNode* result=new TreeNode(nums[medium]); result->left=sortedArrayToBST(leftlist); result->right=sortedArrayToBST(righlist); return result; } };
参考: https://www.tianmaying.com/tutorial/LC108
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