leetcode-109-Convert Sorted List to Binary Search Tree

#include <iostream>
using namespace std;
/*
 题目要求：
 Given a singly linked list where elements are 
 sorted in ascending order, convert it to a height 
 balanced BST.
 这道题可以与108的根据有序数组构建平衡二叉搜索树作对比，由于有序数组可以
 随机访问元素，所以108那道题可以用自顶向下的方法构建树。但是单链表不能随机
 访问，只能顺序访问，根据这个数据结构的特点，自顶向下的建树顺序不太妥当，应当
 用自底向上的建树顺序，从左向右扫描链表，由于比中间数字小的数字都是该数字的左子树，
 所以可以先构建所有可能的右子树，然后构建中间数字的节点，再考虑其右子树。
 */
/**
 * 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:
    TreeNode* sortedListToBST(ListNode* head) {
        int len = 0;
        ListNode* cur = head;
        while (cur) {
            cur = cur->next;
            len++;
        }
        return bottomUp(head, 0, len - 1);
    }
private:
    //注意要用引用参数
    TreeNode* bottomUp(ListNode* &h, int left, int right){
        //如果链表为空，则返回空指针
        if (!h) {
            return NULL;
        }
        if (left > right) {
            return NULL;
        }
        int mid = (left + right) / 2;
        //构建左子树
        TreeNode* lChild = bottomUp(h, left, mid - 1);
        //当前节点
        TreeNode* current = new TreeNode(h->val);
        current->left = lChild;
        h = h->next;
        //构建右子树
        TreeNode* rChild = bottomUp(h, mid + 1, right);
        current->right = rChild;
        return current;
    }
};
int main(int argc, const char * argv[]) {
    ListNode one(1);
    ListNode two(2);
    ListNode three(3);
    one.next = &two;
    two.next = &three;
    Solution s;
    TreeNode *res = s.sortedListToBST(&one);
    return 0;
}