GATE | GATE-CS-2003 | Question 6
Let T(n) be the number of different binary search trees on n distinct elements.
Then , where x is
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.
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.
If n=4 We have 14 possibilities. Taking each number as root and arranging small numbers as left subtree and larger numbers as right subtree.
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.
This solution is contributed by Parul Sharma.
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