MCQs
Total Questions : 12
| Page 1 of 2 pages
Answer: Option B. -> find the median of the set of elements given, make it as root and construct the tree
Sort the given input, find the median element among them, make it as root and construct left and right subtrees with elements lesser and greater than the median element recursively. this ensures the subtrees differ only by height 1.
Answer: Option A. -> public void min(Tree root){
Since all the elements lesser than a given node will be towards the left, iterating to the leftmost leaf of the root will give the minimum element in a binary search tree.
Answer: Option D. -> Undo/Redo operations in a notepad
This is an application of stack.
Question 4. The following lines talks about deleting a node in a binary tree.(the tree property must not be violated after deletion)
i) from root search for the node to be deleted
ii)_____________________________________
iii) delete the node at _____________________
what must be statement ii) and fill up statement iii)
i) from root search for the node to be deleted
ii)_____________________________________
iii) delete the node at _____________________
what must be statement ii) and fill up statement iii)
Answer: Option D. -> ii)-find deepest node,replace with node to be deleted. iii)- delete the deepest node
We just replace a to be deleted node with last leaf node of a tree. this must not be done in case of BST or heaps.
Answer: Option A. -> every node stores data saying which of its children exist in the array
Array cannot represent arbitrary shaped trees it can only be used in case of complete trees hence option a must be done which is one of several ways to use array in such situations.
Question 6. After applying the below operations on a input sequence, what happens?
i. construct a cartesian tree for input sequence
ii. put the root element of above tree in a priority queue
iii. if( priority queue is not empty) then
.search and delete minimum value in priority queue
.add that to output
.add cartesian tree children of above node to priority queue
i. construct a cartesian tree for input sequence
ii. put the root element of above tree in a priority queue
iii. if( priority queue is not empty) then
.search and delete minimum value in priority queue
.add that to output
.add cartesian tree children of above node to priority queue
Answer: Option B. -> sorts the input sequence
The above given steps are for sorting a cartesian tree. cartesian sort is benificial in case of partially sorted set of elements. a cartesian
sort can be considered as a selection or heap sort maintaining a priority queue.
Answer: Option A. -> when there are more insertions or deletions
Though both trees are balanced, when there are more insertions and deletions to make the tree balanced, AVL trees should have more rotations, it would be better to use red-black. but if more search is required AVL trees should be used.
Answer: Option A. -> cache Implementation
Splay trees can be used for faster access to recently accessed items and hence used for cache implementations.
Answer: Option B. -> public void postorder(Tree root){ if (root == null) return;
The left and right children are added first to the stack, followed by the node, which is then popped. Post-order follows LRN policy.
Answer: Option B. -> O(n)
Since you have to go through all the nodes, the complexity becomes O(n).
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