LeetCode 109. Convert Sorted List to Binary Search Tree

Solution1：我的答案
偷鸡摸狗的做法

/*** Definition for singly-linked list.* struct ListNode {*     int val;*     ListNode *next;*     ListNode(int x) : val(x), next(NULL) {}* };*/
/*** Definition for a binary tree node.* 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;vector<int> num;ListNode* cur = head;while (cur) {num.push_back(cur->val);cur = cur->next;}return sortedArrayToBST(num, 0 , num.size() - 1);}TreeNode *sortedArrayToBST(vector<int> &num, int left, int right) {if (left > right) return NULL;int mid = (left + right) / 2;TreeNode *cur = new TreeNode(num[mid]);cur->left = sortedArrayToBST(num, left, mid - 1);cur->right = sortedArrayToBST(num, mid + 1, right);return cur;}
};

Solution2：
正儿八经的做法
【重点】：学习快慢指针找链表中点的代码写法
参考网址：http://www.cnblogs.com/grandyang/p/4295618.html
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：

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;//last是slow的前一个结点，也是隶属于左子树链表的最后一个结点slow = slow->next;fast = fast->next->next;}fast = slow->next;//把fast设置为隶属于右子树链表的起点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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