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。
