[109]有序链表转换二叉搜索树

给定一个单链表，其中的元素按升序排序，将其转换为高度平衡的二叉搜索树。
本题中，一个高度平衡二叉树是指一个二叉树每个节点 的左右两个子树的高度差的绝对值不超过 1。
示例:

给定的有序链表： [-10, -3, 0, 5, 9],
一个可能的答案是：[0, -3, 9, -10, null, 5], 它可以表示下面这个高度平衡二叉搜索树：
      0
     / \
   -3   9
   /   /
 -10  5

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* sortedListToBST(ListNode* head) {
        if(head == nullptr) {
            return nullptr;
        }else if(head->next == nullptr) {
            return new TreeNode(head->val);
        }else if(head->next->next == nullptr) {
            TreeNode* Temphead = new TreeNode(head->next->val);
            Temphead->left = new TreeNode(head->val);
            return Temphead;
        }       
        ListNode* slow, * fast, *pre;
        slow = head;
        fast = head->next;
        pre = nullptr;
        while(fast!= nullptr && fast->next!= nullptr){
            cout<<slow->val<<endl;
            pre = slow;
            slow = slow->next;
            fast = fast->next->next;
        }
        cout<<" midd "<<slow->val<<endl;
        cout<<" pre  "<<pre->val<<endl;
        TreeNode* res = new TreeNode(slow->val);
        ListNode* next = slow->next;
        pre->next = nullptr;
        cout<<head->val<<" "<<next->val<<endl;
        res->left = sortedListToBST(head);
        res->right = sortedListToBST(next);
        return res;
    }
};