Binary Search Trees(B Tree) MCQ question and answer with solution | Data Structure MCQs

A. By counting the number of edges from the root to the node.

B. By counting the number of nodes in the subtree.

C. By finding the height of the tree.

D. By counting the number of children of the node.

A. To implement dynamic sets and lookup tables.

B. To perform sorting operations.

C. To manage hierarchical data.

D. To implement priority queues.

A. Nodes are visited in ascending order.

B. Nodes are visited in descending order.

C. Nodes are visited in the order of their depth.

D. Nodes are visited in the order they are inserted.

A. Insert duplicates into the right subtree or use a counter at each node.

B. Insert duplicates into the left subtree.

C. Store duplicates in a separate data structure.

D. Ignore duplicate values.

A. The number of edges from the node to its deepest descendant.

B. The number of nodes in the subtree.

C. The number of edges from the node to its parent.

D. The total number of nodes in the tree.

A. O(1)

B. O(nlogd)

C. O(logd)

D. O(d)

A. yes because we are overcoming the need of pointers and so space efficiency

B. yes because array values are indexable

C. No it is not efficient in case of sparse trees and remaning cases it is fine

D. No linked list representation of tree is only fine

A. used to store strings efficiently

B. used to store integers efficiently

C. can be used in process schedulers, maps, sets

D. for efficient sorting

A. O(N)

B. O(log N)

C. O( N log N)

D. O(N²)

A.

public void inorder(Tree root)
{
	System.out.println(root.data);
	inorder(root.left);
	inorder(root.right);
}

B.

public void inorder(Tree root)
{
	inorder(root.left);
	System.out.println(root.data);
	inorder(root.right);
}

C.

public void inorder(Tree root)
{
	System.out.println(root.data);
	inorder(root.right);
	inorder(root.left);
}

D.

public void inorder(Tree root)
{
	inorder(root.right);
	inorder(root.left);
	System.out.println(root.data);
}