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:

class Solution {
public:
    TreeNode *sortedListToBST(ListNode *head) {
        if (head == NULL) return NULL;
        ListNode* slow = head, *fast = head, *pre = NULL;
        while(slow != NULL && fast != NULL && fast->next != NULL) {
            pre = slow;
            slow = slow->next;
            fast = fast->next->next;
        }
        
        TreeNode* root = new TreeNode(slow->val);
        
        TreeNode* left = NULL;
        if (pre != NULL) {
            pre->next = NULL;
            left = sortedListToBST(head);
        } 
        
        TreeNode* right = sortedListToBST(slow->next);
        
        root->left = left;
        root->right = right;
        
        return root;
    }
};

bottom up:

class Solution {
public:
    TreeNode *sortedListToBST(ListNode *head) {
        int n = 0; 
        ListNode* p = head;
        while (p != NULL) {
            n++;
            p = p->next;
        }
        return recursive(head, 0, n - 1);
    }
    
    TreeNode* recursive(ListNode* &head, int start, int end) {
        if (head == NULL) return NULL;
        if (start > end) return NULL;
        int mid = start + (end - start) / 2;
        TreeNode* left = recursive(head, start, mid - 1);
        TreeNode* root = new TreeNode(head->val);
        head = head->next;
        TreeNode* right = recursive(head, mid + 1, end);
        root->left = left;
        root->right = right;
        return root;
    }
};

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。