Access time of a binary search tree may go worse in terms of...

MCQ

Access time of a binary search tree may go worse in terms of time complexity upto

?(n2)

?(n log n)

?(n)

?(1)

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What is missing in this logic of finding a path in the tree for a given sum (i.e checking whether there will be a path from roots to leaf nodes with given sum)?

checkSum(struct bin-treenode *root , int sum) :
  if(root==null)
    return sum as 0
  else :
     leftover_sum=sum-root_node-->value
     //missing

code for having recursive calls to either only left tree or right trees or to both subtrees depending on their existence

code for having recursive calls to either only left tree or right trees

code for having recursive calls to either only left tree

code for having recursive calls to either only right trees

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State true of false.i) A node is a parent if it has successor nodes.ii) A node is child node if out degree is one.

True, True

True, False

False, True

False, False

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Which of the following remarks about Tree- indexing are true?

It is an m-ary tree

Successful searches should terminate in leaf nodes

Unsuccessful searches may terminate in leaf nodes level of the tree structure

All of these

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What is direct addressing?

Distinct array position for every possible key

Fewer array positions than keys

Fewer keys than array positions

None of the mentioned

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Why the below pseudo code where x is a value, wt is weight factor and t is root node can

when x>t. element Rotate-with-left-child should take place and vice versa

the logic is incorrect

the condition for rotating children is wrong

insertion cannot be performed in weight balanced trees

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Queues

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Binary Trees

B Trees

Heaps

Hash based structures

Graphs

Mixed

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