GATE CS 2014 Set 1 Q.22 (CS Q.12): Binary Trees, Binary Search Trees and Balanced Trees

Consider a rooted $n$ node binary tree represented using pointers. The best upper bound on the time required to determine the number of subtrees having exactly 4 nodes is $O(n^a \log^b n)$ . Then the value of $a + 10b$ is ___________.

Your answer

Official answer and solution

Official answer: 1
From: Official GATE 2014 answer key

Solution

Analyze the time complexity of counting nodes in subtrees using post-order traversal.

We can compute the size of every subtree in a rooted binary tree using a standard bottom-up post-order traversal.

For any node $u$ :

\text{size}(u) = 1 + \text{size}(\text{left}(u)) + \text{size}(\text{right}(u))

At each node, computing its size takes $O(1)$ time given the sizes of its left and right children.
Checking whether $\text{size}(u) == 4$ and updating a counter takes $O(1)$ time.
The traversal visits each of the $n$ nodes exactly once.

Thus, the total time required is $\Theta(n) = O(n)$ .

Comparing $O(n)$ with the given form $O(n^a \log^b n)$ :

a = 1, \quad b = 0

Evaluating the expression:

a + 10b = 1 + 10(0) = 1

Final answer

Common trap

Recomputing the subtree size from scratch at each node in $O(n)$ time, which would lead to an inferior $O(n^2)$ bound.

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2014 Computer Science and Information Technology (CS) Session 1
Question: 22 of 65, printed as CS Q.12
Section: Computer Science and Information Technology

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