Question

) True or false: Any two (possibly unbalanced) binary search trees containing n elements each can be merged into a single balanced binary search tree in O(n) time.

Answer 1

False,

Two BST containing n elements in each BST can be merged by the following methods with complexities other that O(n):

1. Insert elements of first tree to second(Method 1):

  In this approach we are required to take elements of 1st BST one by one and insert those into the 2nd BST.

Insertion of these elements take Lon(n) time, where n is the size of BST.

Thus time complexity of this method comes out to be Log(n) + Log(n+1) … Log(m+n-1), implies,

Complexity is between mLogn and mLog(m+n-1).

2. Merge Inorder Traversals (Method 2):

1. Do inorder traversal of the first tree.

2. Store the traversal in one temporary array array1[ ] (This step takes O(m) time).
3. Do inorder traversal of second tree.

4. Store the traversal in another temporary array array2[ ] (This step takes O(n) time).
3. Both the arrays array1 and array2 are sorted array. Now merge these two arrays into one array of size (m+n) (This step takes O(m+n) time).
4) Now create a balanced tree from the merged array thereby taking time of O(m+n).

Result:

If we compare both methods we can see that Inorder traversal takes less time than the first method as O(m+n).

It cannot be solved with O(n) time as two minimum steps are storing and merging thus taking time O(m+n) and O(m+n) respectively.