[Leetcode] 108. Convert Sorted Array to Binary Search Tree 解题报告
2017-04-25 10:11
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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;
}
};
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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