Convert Sorted List to Binary Search Tree

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

分析： 根据一个有序链表，得到一个平衡二叉搜索树，主要是根据快慢指针得到链表的中点，然后将链表分成两半

/**
 * 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:
    ListNode* findMiddle(ListNode* start){
        ListNode *low = start, *fast = start;
        ListNode* prelow=nullptr;
        while(fast!=nullptr){
            fast = fast->next;
            if(fast){
                fast = fast->next;
                prelow = low;
                low = low->next;
            }
            
        }
        if(prelow)
            prelow->next =nullptr;//break the list
        return low;
    }
    
    TreeNode* sortedListToBST(ListNode* head) {
        if(head==nullptr)
            return nullptr;
        if(head->next ==nullptr)
            return new TreeNode(head->val);
        ListNode* mid =  findMiddle(head);
        TreeNode* root = new TreeNode(mid->val);
        root->left = sortedListToBST(head);
        root->right = sortedListToBST(mid->next);
        return root;
    }
};