[LeetCode] 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.

有序链表转为平衡的二叉搜索树

与数组类似，找到链表的中间节点，链表左半部分递归转化，右半部分递归转化

/**
 * 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) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        if(head == NULL) return NULL;
        if(head->next == NULL) return new TreeNode(head->val);
        ListNode *middle,*pre;
        findMiddle(head,middle,pre);
        pre->next = NULL;
        TreeNode *root = new TreeNode(middle->val);
        root->left = sortedListToBST(head);
        root->right = sortedListToBST(middle->next);
        return root;
    }
    
    void findMiddle(ListNode *head, ListNode *&middle, ListNode *&pre) {
        int len = countLen(head);
        pre = head;
        middle = head->next;
        for(int i = 1; i < len/2; i++) {
            pre = pre->next;
            middle = middle->next;
        }
    }
    
    int countLen(ListNode *head) {
        if(head == NULL) return 0;
        int len = 0;
        while(head != NULL) {
            len++;
            head = head->next;
        }
        return len;
    }
};