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GATE | GATE-CS-2003 | Question 6

  • Difficulty Level : Hard
  • Last Updated : 28 Jun, 2021

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 −

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  • 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.

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