MCQs Library

Traversing A Heap In Order Provides Access To The Data In Numeric Or Alphabetical Order

Removing The Item At The Top Provides Immediate Access To The Key Value With Highest (Or Lowest) Priority

Inserting An Item Is Always Done At The End Of The Array But Requires Maintaining The Heap Property

A Heap May Be Stored In An Array

A Find(X) On Element X Is Performed By Returning Exactly The Same Node That Is Found

A Find(X) On Element X Is Performed By Returning The Root Of The Tree Containing X

A Find(X) On Element X Is Performed By Returning The Whole Tree Itself Containing X

A Find(X) On Element X Is Performed By Returning True

To Perform Union Of Two Sets We Merge The Two Trees By Making The Root Of One Tree Point To The Root Of The Other

To Perform Union Of Two Sets We Merge The Two Trees By Making The Leaf Node Of One Tree Point To The Root Of The Other

Perform Union Of Two Sets Merging Operation Of Trees In Not Required At All

None Of The Given Options

A Threaded Binary Tree Is A Binary Tree In Which Every Node That Does Not Have A Left Child Has A Thread (In Actual Sense A Link) To Its Inorder Successor

A Threaded Binary Tree Is A Binary Tree In Which Every Node That Does Not Have A Right Child Has A Thread (In Actual Sense A Link) To Its Preoredr Successor

A Threaded Binary Tree Is A Binary Tree In Which Every Node That Does Not Have A Right Child Has A Thread (In Actual Sense A Link) To Its Inorder Successor

A Threaded Binary Tree Is A Binary Tree In Which Every Node That Does Not Have A Right Child Has A Thread (In Actual Sense A Link) To Its Postorder Successor

It Is Not A Requirement That A Find Operation Returns Any Specific Name Just That Finds On Two Elements Return The Same Answer If And Only If They Are In The Same Set

One Idea Might Be To Use A Tree To Represent Each Set Since Each Element In A Tree Has The Same Root Thus The Root Can Be Used To Name The Set

Initially Each Set Contains One Element

Initially Each Set Contains One Element And It Does Not Make Sense To Make A Tree Of One Node Only

It Is Not A Requirement That A Find Operation Returns Any Specific Name Just That Finds On Two Elements Return The Same Answer If And Only If They Are In The Same Set

One Idea Might Be To Use A Tree To Represent Each Set Since Each Element In A Tree Has The Same Root Thus The Root Can Be Used To Name The Set

Initially Each Set Contains One Element

Initially Each Set Contains One Element And It Does Not Make Sense To Make A Tree Of One Node Only

Right Thread Of The Right-Most Node Points To The Dummy Node

Left Thread Of The Left-Most Node Points To The Dummy Node

The Left Pointer Of Dummy Node Points To The Root Node Of The Tree

Left Thread Of The Right-Most Node Points To The Dummy Node

This Dummy Node Never Has A Value

This Dummy Node Has Always Some Dummy Value

This Dummy Node Has Either No Value Or Some Dummy Value

This Dummy Node Has Always Some Integer Value

A Binary Tree With N Internal Nodes Has N+1 Internal Links

A Binary Tree With N External Nodes Has 2N Internal Nodes

A Binary Tree With N Internal Nodes Has N+1 External Nodes

None Of Above Statement Is A Property Of The Binary Tree

If (Alpha < Beta)

If (Alpha.Lessthan(Beta))

If (Lessthan(Alpha Beta))

If (Lessthan(Alpha).Beta)

Maintain Sizes (Number Of Nodes) Of All Trees And During Union

Make Smaller Tree The Subtree Of The Larger One

Make The Larger Tree The Subtree Of The Smaller One

Implementation: For Each Root Node I Instead Of Setting Parent[I] To -1 Set It To -K If Tree Rooted At I Has K Nodes

Maintain Sizes (Number Of Nodes) Of All Trees And During Union

Make Smaller Tree The Subtree Of The Larger One

Make The Larger Tree The Subtree Of The Smaller One

Implementation: For Each Root Node I Instead Of Setting Parent[I] To -1 Set It To -K If Tree Rooted At I Has K Nodes

Stack

Queue

Link List

All Of These

Bubble Sort

Insertion Sort

Quick Sort

Merge Sort

None Of The Given Options

Call By Passing The Value Of The Argument

Call By Passing Reference Of The Argument

Call By Passing The Address Of The Argument

They Always Point To?Different Memory Locations

They May Point To A Single?Memory Location

The Address Of Two Pointer Variables Is Same

None Of These

Linked List

Stack

Queue

Tree

Electrical Connectivity

Set Of People

<= Relation

Set Of Pixels

Reflexive

Symmetric

Transitive

Associative

Each List Si Contains The Special Keys + Infinity And - Infinity

List S0 Contains The Keys Of S In Non-Decreasing Order

Each List Is A Subsequence Of The Previous One

List Sh Contains Only The N Special Keys