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,...
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 ofeverynode never differ by more than 1. Example: AI检测代码解析 Given th...
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
TreeNode*sortedArrayToBST(vector<int> &num) {if(num.empty())returnNULL; TreeNode*root=NULL;inti;for(i=0; i<(int)num.size(); i++) insert(root,num[i]);returnroot; }voidinsert(TreeNode *&root,intkey) {intlhigh,rhigh;if(root==NULL) ...
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...
The suffix binary search tree[Irving and Love, JDA'03] is a dynamic data structure that can answer this problem dynamically in the sense that insertions and deletions of positions in P are allowed. While a standard binary search tree on strings needs to store two longest-common prefix (LCP)...
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...
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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;}