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;}