leetcode刷题，总结，记录，备忘109

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leetcode109

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.

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使用二分法，然后分别分为2个分支，进行递归操作，还是比较简单的，但是貌似算法耗时有点多，40ms，，但是在leetcode提交通过的解决方案中排的比较靠后。。

/** * 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 * result;        void funciton(vector<int> vi, TreeNode * root)    {        if (root == NULL)        {            root = new TreeNode(0);            result = root;        }                if (vi.size() == 1)        {            root->val = vi[0];            root->left = NULL;            root->right = NULL;        }        else if (vi.size() == 2)        {            root->val = vi[0];            root->left = NULL;            root->right = new TreeNode(vi[1]);        }        else        {            int mid = vi.size() / 2;            root->val = vi[mid];            root->left = new TreeNode(0);            root->right = new TreeNode(0);            vector<int> vl(vi.begin(), vi.begin() + mid);            vector<int> vr(vi.begin() + mid + 1, vi.end());            funciton(vl, root->left);            funciton(vr, root->right);        }    }        TreeNode* sortedListToBST(ListNode* head) {        vector<int> vi;        if (head == NULL)        {            return NULL;        }                while (head)        {            vi.push_back(head->val);            head = head->next;        }                funciton(vi, NULL);                return result;    }};

然后在讨论区看到一个解法，受到的启发，还记得leetcode之前有个判断1个链表是否是循环的题，使用快慢指针，这个方法就是使用快慢指针，找到中间的指针，然后也是1分为2，递归做二分法，比较巧妙的做法，感觉比我自己想的解法更好，下面也把这个解决方法贴出来。

/** * 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) {        if (head == NULL)        {            return NULL;        }                ListNode * fast = head;        ListNode * slow = head;        ListNode * pre = NULL;                while (fast && fast->next)        {            fast = fast->next->next;            pre = slow;            slow = slow->next;        }                if (pre)        {            pre->next = NULL;        }        else        {            head = NULL;        }                TreeNode * root = new TreeNode(slow->val);        root->left = sortedListToBST(head);        root->right = sortedListToBST(slow->next);                        return root;    }};

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