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.

top-down还是o(nlgn)。自己没有细想，不过查了一下，发现还是有o(n)的，用的是bottom-up. 

The bottom-up approach enables us to access the list in its order while creating nodes. Each time you are stucked with the top-down approach, give bottom-up a try. Although bottom-up approach is not the most natural way we think, it is extremely helpful in some cases. However, you should prefer top-down instead of bottom-up in general, since the latter is more difficult to verify in correctness.

top down:

 1 class Solution {

 2 public:

 3     TreeNode *sortedListToBST(ListNode *head) {

 4         if (head == NULL) return NULL;

 5         ListNode* slow = head, *fast = head, *pre = NULL;

 6         while(slow != NULL && fast != NULL && fast->next != NULL) {

 7             pre = slow;

 8             slow = slow->next;

 9             fast = fast->next->next;

10         }

11         

12         TreeNode* root = new TreeNode(slow->val);

13         

14         TreeNode* left = NULL;

15         if (pre != NULL) {

16             pre->next = NULL;

17             left = sortedListToBST(head);

18         } 

19         

20         TreeNode* right = sortedListToBST(slow->next);

21         

22         root->left = left;

23         root->right = right;

24         

25         return root;

26     }

27 };

bottom up:

 1 class Solution {

 2 public:

 3     TreeNode *sortedListToBST(ListNode *head) {

 4         int n = 0; 

 5         ListNode* p = head;

 6         while (p != NULL) {

 7             n++;

 8             p = p->next;

 9         }

10         return recursive(head, 0, n - 1);

11     }

12     

13     TreeNode* recursive(ListNode* &head, int start, int end) {

14         if (head == NULL) return NULL;

15         if (start > end) return NULL;

16         int mid = start + (end - start) / 2;

17         TreeNode* left = recursive(head, start, mid - 1);

18         TreeNode* root = new TreeNode(head->val);

19         head = head->next;

20         TreeNode* right = recursive(head, mid + 1, end);

21         root->left = left;

22         root->right = right;

23         return root;

24     }

25 };

首先是要得到整个list的长度，有了这个长度才能知道最终的BST是什么样子的。然后是先用左边的元素完成左边的子树构建，要确保构建完之后，list的head指针必须更新到末尾。所以这里需要传引用,ListNode*&。整个算法复杂度为o(n)。

另一种做法：

既然知道长度之后就知道最终的BST是什么样子的，那么可以先构建一颗空树。然后inorder traversal填进去就可以。