GATE | GATE-CS-2003 | Question 6

Let T(n) be the number of different binary search trees on n distinct elements.
Then GATECS2003Q7, where x is
(A) n-k+1
(B) n-k
(C) n-k-1
(D) n-k-2


Answer: (B)

Explanation: The idea is to make a key root, put (k-1) keys in one subtree and remaining n-k keys in other subtree.

A Binary Search Tree (BST) is a tree in which all the nodes follow the below-mentioned properties −

  • The left sub-tree of a node has a key less than or equal to its parent node’s key.
  • The right sub-tree of a node has a key greater than to its parent node’s key.

Now construction binary search trees from n distinct number-
Lets for simplicity consider n distinct numbers as first n natural numbers (starting from 1)
If n=1 We have only one possibility, therefore only 1 BST.
If n=2 We have 2 possibilities , when smaller number is root and bigger number is the right child or second when the bigger number is root and smaller number as left child.

 

parul_1



If n=3 We have 5 possibilities. Keeping each number first as root and then arranging the remaining 2 numbers as in case of n=2.

parul_2

 

If n=4 We have 14 possibilities. Taking each number as root and arranging smaal numbers as left subtree and larger numbers as right subtree.
parul_4

Thus we can conclude that with n distinct numbers, if we take ‘k’ as root then all the numbers smaller than k will left subtree and numbers larger than k will be right subtree where the the right subtree and left subtree will again be constructed recursively like the root.
Therefore,

parul5

 

This solution is contributed by Parul Sharma.

Quiz of this Question

My Personal Notes arrow_drop_up
Article Tags :

Be the First to upvote.


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.