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

這道題是要求把有序連結串列轉為二叉搜尋樹，和之前那道Convert Sorted Array to Binary Search Tree 將有序陣列轉為二叉搜尋樹思路完全一樣，只不過是操作的資料型別有所差別，一個是陣列，一個是連結串列。陣列方便就方便在可以通過index直接訪問任意一個元素，而連結串列不行。由於二分查詢法每次需要找到中點，而連結串列的查詢中間點可以通過快慢指標來操作，可參見之前的兩篇部落格

/**
 * 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;
    }
};