【LeetCode】Convert Sorted List to Binary Search Tree

/**
 * 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 == NULL)
        {//空
            return NULL;
        }
        else if(head->next == NULL)
        {//一个节点
            TreeNode* root = new TreeNode(head->val);
            return root;
        }
        else
        {
            ListNode* mid = findMid(head);  //找出中间节点作为根节点（满足保持平衡的要求）
            ListNode* lefthead = head;      //左子串头节点为原链表头节点
            ListNode* righthead = mid->next;//右子串头节点为中间节点的下一个节点
            TreeNode* root = new TreeNode(mid->val);    //递归下去即可
            root->left = sortedListToBST(lefthead);
            root->right = sortedListToBST(righthead);
            return root;
        }
    }

    ListNode* findMid(ListNode* head)
    {//调用前提是head != NULL
     //若链表为偶数，则返回均分后半子串的指针；若链表为奇数，则返回中间节点的指针
        ListNode* pre = NULL;
        ListNode* fast = head;
        ListNode* slow = head;
        while(fast != NULL)
        {
            fast = fast->next;
            if(fast != NULL)
            {
                fast = fast->next;
                pre = slow;
                slow = slow->next;
            }
        }
        pre->next = NULL;   //前子串尾部置空
        return slow;
    }
};