aai 2018

Question

 

Consider an AVL tree constructed by inserting the elements 18, 10,5,3,27,7,35,43,32 one by one . Which of following options gives the number of comparisions to search the key value 45 in this AVL tree?

• A. 3
• B. 5
• C. 4
• D. 2 

in this question ans 5 is correct or 4 ?

i think ans  5 is correct 4 for non match element one for not present so ans will be 5

Comments

$4$ is correct.

Answer 1

see this

final $AVL$ tree look like this

 

Comments

i already commented on another question of yours,

first edit the question with the text... it is your responsibility !

https://gateoverflow.in/blog/5823/screenshot

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ok sir

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in the trees we have to take care of links, i mean before accessing it's value, first we have to check it is null or not.

int search(struct node *current, int element)

{

     if( current -> data != element )              //  main comparission

     {

        if( current -> next != element )    // actually we have to consider this comparission also

            return search(current -> Next , element );

       else

          return -1;

    }

   else

      return 1;

}

if the function returns 1, then element found, otherwise element not found.

options are not matching... So we can understand questioner interested in main comparissions only.

But in the Hashing we can't check like that, So one more comparission needed !