#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) {...
Main=Solution() 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...
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]...
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]...
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; ...
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 -- 5:37 App Leetcode-0033. Search in Rotated Sorted Array 5 -- 2:23 App Leetcode-0026. Remove Duplicates from Sorted Array ...
Convert Sorted Array to Binary Search Tree Given an array where elements are sorted in ascending order, convert it to a height balanced BST. 解题思路: 题意为构建有序数组的二分查找树。 比較简单。用递归方法就可以,中间的元素作为根节点,前半部分作为左孩子树,右半部分作为右孩子树。
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...
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...
This is accomplished by generating a balanced red-black binary search tree for a sorted array of data items wherein the information stored at each node of the tree includes: a key value; a color; a left pointer to indicate the presence of a child to the left of the node; a right ...