[Leetcode] 108. Convert Sorted Array to Binary Search Tree 解题报告
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题目:
Given an array where elements are sorted in ascending order, convert it to a height balanced BST.
思路:
也是很简单的一道题目:首先选取数组的中位数作为根节点,然后递归地构造左子树和右子树,最后合成并返回结果。算法的时间复杂度是O(n),空间复杂度是O(1),因为没有开辟额外的空间。但是如果算上递归调用所占用的栈空间,则空间复杂度为O(logn)。
代码:
/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public: TreeNode* sortedArrayToBST(vector<int>& nums) { return sortedArrayToBST(nums, 0, nums.size()); }private: TreeNode* sortedArrayToBST(vector<int>& nums, int start, int len) { if (len <= 0) { return NULL; } int mid = start + len / 2; int len1 = mid - start; int len2 = len - len1 - 1; TreeNode* root = new TreeNode(nums[mid]); root->left = sortedArrayToBST(nums, start, len1); root->right = sortedArrayToBST(nums, mid + 1, len2); return root; }};
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