Home > Recursion, Tree > Find ‘ceil’ of a key in a Binary Search Tree (BST)

Find ‘ceil’ of a key in a Binary Search Tree (BST)

Long long time, eh?! But it has been a busy time at work. 🙂

So lets solve this problem. An example tree follows:

treeNow lets say you have the following queries –

q = 13, Answer = Does Not Exist

q = 5, Answer = 6

How do you go about finding the ceil?

One way to do it is Inorder Traversal of the tree. This will take O(n) time and then get you kicked out of the interview. Did you not read “BST” ? So we should be using that particular property of the tree!

Here is the algorithm you should be following in short

“Start with the root as current node.

If value(current node) == key, you are done!

If value(current node) > key, make answer = value(current node) and current node = leftNode(current node).

Else if value(current node)<key, make current node = rightNode(currentNode).

You have your answer once you exhaust nodes in the path you are moving on this fashion. If no value was assigned to answer, this indicates you kept moving right throughout and the tree has no value >= key.”

That does not sound right? Well, try it out for yourself! And the complexity? O(log n). Whoopie!

Advertisements
Categories: Recursion, Tree Tags: ,
  1. Nikos Karamitrou
    November 27, 2014 at 4:50 am

    Actually the complexity is not O(logn). It’s O(height of the tree) where height of the tree could easily be “n”.

  1. No trackbacks yet.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: