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Sunday, January 27, 2013

[LeetCode] Convert Sorted List to Binary Search Tree, Solution


Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.
» Solve this problem

[Thoughts]
It is similar with "Convert Sorted Array to Binary Search Tree". But the difference here is we have no way to random access item in O(1).

If we build BST from array, we can build it from top to bottom, like
1. choose the middle one as root,
2. build left sub BST
3. build right sub BST
4. do this recursively.

But for linked list, we can't do that because Top-To-Bottom are heavily relied on the index operation.
There is a smart solution to provide an Bottom-TO-Top as an alternative way, http://leetcode.com/2010/11/convert-sorted-list-to-balanced-binary.html

With this, we can insert nodes following the list’s order. So, we no longer need to find the middle element, as we are able to traverse the list while inserting nodes to the tree.

[Code]
1:    TreeNode *sortedListToBST(ListNode *head) {  
2:      // Start typing your C/C++ solution below  
3:      // DO NOT write int main() function  
4:      int len =0;  
5:      ListNode *p = head;  
6:      while(p)  
7:      {  
8:        len++;  
9:        p = p->next;  
10:      }  
11:      return BuildBST(head, 0, len-1);  
12:    }  
13:    TreeNode* BuildBST(ListNode*& list, int start, int end)  
14:    {  
15:      if (start > end) return NULL;  
16:      int mid = (start+end)/2;   //if use start + (end - start) >> 1, test case will break, strange!
17:      TreeNode *leftChild = BuildBST(list, start, mid-1);  
18:      TreeNode *parent = new TreeNode(list->val);  
19:      parent->left = leftChild;  
20:      list = list->next;  
21:      parent->right = BuildBST(list, mid+1, end);  
22:      return parent;  
23:    }  


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