#include<iostream>#include<new>#include<vector>#include<cmath>usingnamespacestd;//Definition for binary treestructTreeNode {intval; TreeNode*left; TreeNode*right; TreeNode(intx) : val(x), left(NULL), right(NULL) {} };classSolution {public: TreeNode*sortedArrayToBST(vector<int> &num) {...
Can you solve this real interview question? Convert Sorted Array to Binary Search Tree - Given an integer array nums where the elements are sorted in ascending order, convert it to a height-balanced binary search tree. Example 1: [https://assets.lee
Main.sortedArrayToBST(num1) 验证结果: Runtime:72 ms, faster than24.34%ofPython3online submissions forConvert Sorted Array to Binary Search Tree. Memory Usage:15.4 MB, less than5.70%ofPython3online submissions forConvert Sorted Array to Binary Search Tree. 速度太慢了,看来本题优化的空间还很大,...
For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees ofeverynode never differ by more than 1. Example: AI检测代码解析 Given the sorted array: [-10,-3,0,5,9], One possible answer is: [0,-3,9,-10,null,5], which ...
Given an array where elements are sorted in ascending order, convert it to a height balanced BST. 数组的中点肯定就是平衡BST的根节点,以此规律递归。 /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; ...
public: TreeNode *sortedArrayToBST(vector<int> &num) { return create(num,0,num.size()-1); } TreeNode* create(vector<int>num,int L,int R) { if(L>R)return NULL; int mid=(R+L+1)/2; //BST树总是从左到右排满的,如果不优先选右边的这个,{1,3}结果为{1,#,3},而实际结果应为...
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.Example:1.一看之下感觉和108很像，后来发现是linklist，就又不会做了=。=于是在youtube上找到了公瑾讲解 2.这道题...
A binary search tree is structured so that keys or addresses associated with data in the bottom vertices are arranged in a predetermined order, such as ascending key address order. The root vertex and each hierarchy vertex contains the lowest value key from each child vertex and are thus ...
解析 高度平衡,保险起见,左右节点个数相同且均衡分配即可。二叉树问题还是递归问题,将数组分割成左右两个部分,然后递归求解即可。 伪代码 root=mid(nums)root.left=f(nums[:half])root.Right=f(nums[half+1:]) 代码 funcsortedArrayToBST(nums[]int)*TreeNode{iflen(nums)==0{returnnil}k:=len(nums)/2...
intstart,intend){if(start==end){returnnull;}intmid=(start+end)>>>1;TreeNoderoot=newTreeNode(nums[mid]);root.left=sortedArrayToBST(nums,start,mid);root.right=sortedArrayToBST(nums,mid+1,end);returnroot;}