#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) {...
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]...
root.left=BuildTree(nums,s,middle-1); root.right=BuildTree(nums,middle+1,e);returnroot; }
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]...
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: Given the sorted array: [-10,-3,0,5,9], One possible answer is: [0,-3,9,-10,null,5], which represents the fo...
* Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: TreeNode* sortedArrayToBST(vector<int>& nums) { ...
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...
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)...
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{3,1} TreeNode*root=new TreeNode(num[mid]); ...
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)/2node:...