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]...
LeetCode 33 Search in Rotated Sorted Array [binary search] 给出排序好的一维无重复元素的数组,随机取一个位置断开,把前半部分接到后半部分后面,得到一个新数组,在新数组中查找给定数的下标,如果没有,返回 1。时间复杂度限制$O(log_2n)$
1publicTreeNode sortedArrayToBST(int[] num) {2if(num.length == 0)returnnull;3elseif(num.length == 1)returnnewTreeNode(num[0]);4else{5TreeNode root = sortedArrayToBST(num, 0, num.length-1);6returnroot;7}8}9publicTreeNode sortedArrayToBST(int[] num,intstart,intend) {10if(end ...
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. 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. ...
Convert Sorted Array to Binary Search Tree Given an array where elements are sorted in ascending order, convert it to a height balanced BST. 本题是二分法的高级应用,每次选择root都是从数列的中间选择,那么最后构造出来的二叉树一定是高度平衡的BST了。
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 ...
a height balanced BST BST的中序遍历是一个sorted-array,再构造回去成一个BST,先将中间的元素作为根节点,这个节点的左右分别是左子树和右子树。如此递归地...
Leetcode: Convert Sorted Array to Binary Search Tree,Givenanarraywhereelementsaresortedinascendingorder,convertittoaheightbalancedBST.基本上一次过,要注意边界条件的问题:如果在recursion里有两个参数intbegin,end,递...
LeetCodeの制約ではlen(nums) >= 1が保証されてますが、空配列だとIndexErrorになりますね。 👍 1 arai60/21-29_Tree_BT_BST/24_108_Convert Sorted Array to Binary Search Tree/level_3.py def helper(left, right): if left > right: return None mid = (left + right) // 2 hay...