[LeetCode] Convert Sorted List to Binary Search Tree 将有序链表转为二叉搜索树

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;
}
};

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