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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可以通过快慢指针来操作,可参见之前的两篇博客Reorder List 链表重排序和Linked List Cycle II 单链表中的环之二有关快慢指针的应用。找到中点后,要以中点的值建立一个数的根节点,然后需要把原链表断开,分为前后两个链表,都不能包含原中节点,然后再分别对这两个链表递归调用原函数,分别连上左右子节点即可。代码如下:
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ /** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: TreeNode *sortedListToBST(ListNode *head) { if (!head) return NULL; if (!head->next) return new TreeNode(head->val); ListNode *slow = head; ListNode *fast = head; ListNode *last = slow; while (fast->next && fast->next->next) { last = slow; slow = slow->next; fast = fast->next->next; } fast = slow->next; last->next = NULL; TreeNode *cur = new TreeNode(slow->val); if (head != slow) cur->left = sortedListToBST(head); cur->right = sortedListToBST(fast); return cur; } };
[LeetCode] Convert Sorted List to Binary Search Tree 将有序链表转为二叉搜索树
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原文地址:http://www.cnblogs.com/grandyang/p/4295618.html