Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST. Example 2 1->2->3 => / \ 1 3 分析:非常简单,用递归即可.需要注意返回mid node的时候,要把整个list分成两半. /** * Definition for singly-linked list. * public class ListN…

题目: Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST. Example 2 1->2->3 => / \ 1 3   题解: Solution 1 () class Solution { public: TreeNode *sortedListToBST(ListNode *head) { if (!head) ret…

(http://leetcode.com/2010/11/convert-sorted-list-to-balanced-binary.html) Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST. Code: BinaryTree* sortedListToBST(ListNode *& list, int start, int…

(http://leetcode.com/2010/11/convert-sorted-array-into-balanced.html) Given an array where elements are sorted in ascending order, convert it to a height balanced BST. Code: BinaryTree* sortedArrayToBST(int arr[], int start, int end) { if (start > en…

题目:将非递减有序的链表转化为平衡二叉查找树! 参考的博客:http://blog.csdn.net/worldwindjp/article/details/39722643 利用递归思想:首先找到链表的中间节点,于是链表被分为了由该中间节点划分开来的两部分.递归地处理这两部分,最终便得到了平衡二叉查找树. /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * Lis…

Given an array where elements are sorted in ascending order, convert it to a height balanced BST.思路:对于树来说,自顶向下的递归是很好控制的,自底向上的递归就很容易让脑神经打结了.本来想仿照排序二叉树的中序遍历,逆向地由数组构造树,后来发现这样做需要先确定树的结构,不然会一直递归下去,不知道左树何时停止.代码: TreeNode *addNode(vector<int> &num, int…

Convert Sorted List to Binary Search Tree Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST. 为了满足平衡要求,容易想到提出中间节点作为树根,因为已排序,所以左右两侧天然满足BST的要求. 左右子串分别递归下去,上层根节点连接下层根节点即可完成. 递归找中点,然后断开前后两段链表,并继续找…

108. Convert Sorted Array to Binary Search Tree 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 tw…

/* * 109.Convert sorted list to BST * 2016.12.24 by Mingyang * 这里的问题是对于一个链表我们是不能常量时间访问它的中间元素的. * 这时候就要利用到树的中序遍历了,按照递归中序遍历的顺序对链表结点一个个进行访问,而我们要构造的二分查找树正是按照链表的顺序来的. * 思路就是先对左子树进行递归 * 然后将当前结点作为根,迭代到下一个链表结点,最后在递归求出右子树即可.整体过程就是一次中序遍历,时间复杂度是O(n),空间复杂度是栈空间O(…

Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST. 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…