#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) {...
TreeNode *root=new TreeNode(num[mid]); if(begin<mid){ root->left=constructtree(num,begin,mid-1); root->right=constructtree(num,mid+1,end); } return root; } TreeNode *sortedArrayToBST(vector<int> &num) { if(num.size()==0) return nullptr; return constructtree(num,0,num.size()...
Given an integer arraynumswhere the elements are sorted inascending order, convertit to aheight-balancedbinary search tree. Example 1: Input:nums = [-10,-3,0,5,9]Output:[0,-3,9,-10,null,5]Explanation:[0,-10,5,null,-3,null,9] is also accepted: Example 2: Input:nums = [1,3]...
public TreeNode sortedArrayToBST(int[] nums) { return sortedArrayToBST(nums, 0, nums.length); } private TreeNode sortedArrayToBST(int[] nums, int start, int end) { if (start == end) { return null; } int mid = (start + end) >>> 1; TreeNode root = new TreeNode(nums[mid]...
Binary Search Tree Iterator 74 -- 1:31 App Leetcode-0081. Search in Rotated Sorted Array II 84 -- 2:53 App Leetcode-0144. Binary Tree Preorder Traversal 63 -- 4:35 App Leetcode-0101. Symmetric Tree 7 -- 1:29 App Leetcode-0167. Two Sum II - Input Array Is Sorted 12...
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; ...
题目链接:https://leetcode.com/problems/convert-sorted-array-to-binary-search-tree/ 题目: Given an array where elements are sorted in ascending order, convert it to a height balanced BST. 思路: 新建一个结点保存mid值,该结点的左右子树也递归生成,这是个常用的模板 ...
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...
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...
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)/2node:=&TreeNode{Val:nums[k]}node.Left=sortedArrayToBST(nums[:k])node.Right=sortedArrayToBST(nums[k+1:])returnnode}...