Given an integer array nums where the elements are sorted in ascending order, convert it to a height-balanced binary search tree.
A height-balanced binary tree is a binary tree in which the depth of the two subtrees of every node never differs by more than one.
给定一个内部元素按照升序排列的数组,请将其转化成高度平衡的二叉搜索树。
输入 | 输出 |
---|---|
nums = [-10,-3,0,5,9] | [0,-3,9,-10,null,5] |
nums = [1,3] | [3,1] |
IDE:VS2019
IDE版本:16.10.1
语言:c++11
#include "Solution.h" #include <vector> // std::vector #include<iostream> // std::cout using namespace std; // 主程序 int main() { // 输入 vector<int> nums= { -10, -3, 0, 5, 9 }; Solution solution; // 实例化Solution TreeNode* output = solution.sortedArrayToBST(nums); // 主功能 }
#pragma once #include<vector> // std::vector #include<algorithm> using namespace std; //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) {} }; //主功能 class Solution { public: TreeNode* sortedArrayToBST(vector<int>& nums) { if (nums.size() == 0) return nullptr; // 空数组情况 int mid = nums.size() / 2; // 定义中点值 TreeNode* node = new TreeNode(nums[mid]); auto leftTree = vector<int>(nums.begin(), nums.begin() + mid); // 左边树结构 auto rightTree = vector<int>(nums.begin() + mid + 1, nums.end()); // 右边树结构 if (mid != 0) // 左边树结构递归 node->left = sortedArrayToBST(leftTree); // 递归 if (mid != nums.size() - 1) node->right = sortedArrayToBST(rightTree); return node; } };
参考:https://blog.csdn.net/u012814856/article/details/77894863
参考:
https://blog.csdn.net/u012814856/article/details/77894863
最主要的作用就是封装。封装的好处就是可以再次利用。让使用者不必关心这个是什么,只要根据定义使用就可以了。
class中默认的成员访问权限是private的,而struct中则是public的。 (2)class继承默认是private继承,而从struct继承默认是public继承。
参考:https://blog.csdn.net/qq_33973359/article/details/105511966
参考:https://www.cnblogs.com/qyaizs/articles/2039101.html
参考:https://www.cnblogs.com/zhengfa-af/p/8144786.html