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[Leetcode]108. Convert Sorted Array to Binary Search Tree (C++)

题目描述

题目链接:108. Convert Sorted Array to Binary Search Tree

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.

例子

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

解题思路

由于已经给定了一个排序数组,所以构建高度平衡二叉搜索树比较简单,只需要取中间值作为根结点,对于两侧子数组进行递归操作,返回的根结点分别作为左右子树即可,代码如下:

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/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left),
* right(right) {}
* };
*/
#include <vector>

class Solution {
public:
TreeNode* sortedArrayToBST(vector<int>& nums) {
return buildTree(nums, 0, nums.size() - 1);
}

private:
TreeNode* buildTree(std::vector<int>& nums, int begin, int end) {
if (begin > end) {
return nullptr;
}

int mid = (begin + end) / 2;
TreeNode* root = new TreeNode(nums[mid]);
root->left = buildTree(nums, begin, mid - 1);
root->right = buildTree(nums, mid + 1, end);
return root;
}
};
  • 时间复杂度: O(n)
  • 空间复杂度: O(n) — n 个节点作为新的二叉搜索树

GitHub 代码同步地址: 108.ConvertSortedArrayToBinarySearchTree.cpp

其他题目:
GitHub: Leetcode-C++-Solution
博客: Leetcode-Solutions