1.题目描述 Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST. 2.解法分析 题意为将一个按升序排列的单链表转换为一个平衡的BST,将单链表转换为平衡树很容易,然后按照先序遍历赋值即可。时间复杂度和空间复杂度都是O(N),时间复杂度上没什么好...
https://leetcode.com/problems/convert-sorted-list-to-binary-search-tree/ 题目: Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST. 思路: 简单起见,先把链表值存到数组中,用数组递归感觉会简单点。。 算法: public int getListLength(ListNod...
* public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * }*/publicclassSolution {publicTreeNode sortedListToBST(ListNode head) {returnrec(head,null); }publicTreeNode rec(ListNode start, ListNode end) {if(start ==end) {returnnull...
Can you solve this real interview question? Convert to Base -2 - Given an integer n, return a binary string representing its representation in base -2. Note that the returned string should not have leading zeros unless the string is "0". Example 1:
23 -- 6:11 App Leetcode-0114. Flatten Binary Tree to Linked List 89 -- 1:36 App Leetcode-0067. Add Binary 75 -- 3:26 App Leetcode-0102. Binary Tree Level Order Traversal 43 -- 4:43 App Leetcode-0095. Unique Binary Search Trees II 82 -- 6:19 App Leetcode-0096. Uniq...
Can you solve this real interview question? Convert Binary Number in a Linked List to Integer - Given head which is a reference node to a singly-linked list. The value of each node in the linked list is either 0 or 1. The linked list holds the binary rep
基本思路同 LeetCode: 108. Convert Sorted Array to Binary Search Tree 题解。由于给定数据结构是链表,因此只能先生成左子树,以使迭代指针指向链表中根节点的元素。 AC 代码 /** * Definition for singly-linked list. * struct ListNode { ...
Leetcode - Convert Sorted Array to Binary Search Tree Paste_Image.png My code: /** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; }...
reference: https://discuss.leetcode.com/topic/62880/java-straightforward-solution-ten-binary-hex 看了答案会做的。记住,用 base = 1 进行逐位比较的时候,判断条件是: ** (num & base) != 0 ** Anyway, Good luck, Richardo! -- 10/14/2016...
这道题是要求把有序链表转为二叉搜索树,和之前那道Convert Sorted Array to Binary Search Tree 将有序数组转为二叉搜索树思路完全一样,只不过是操作的数据类型有所差别,一个是数组,一个是链表。数组方便就方便在可以通过index直接访问任意一个元素,而链表不行。由于二分查找法每次需要找到中点,而链表的查找中间点...