MadeEasy Subject Test: Programming & DS - Binary Tree

Question

Consider the following program:

void find(struct Node *node)

{

struct Node *ptr,*q;

q = (struct Node *)malloc(sizeof(struct Node));

q->left = NULL;

q->right = NULL;

if(node == NULL) return;

find(nod->left);

find(nod->right);

ptr = node->left;

q->value = node->value;

node->left = q;

node->left->left = ptr;

}

If the root of the following tree is passed to the above function, by main function the sum of all the keys in the resultant tree produced by find() is given by___?

Comments

Yes, it is a wrong question, I miss read it. sorry for that still my focus was on tracing the program. The question was not having an option as compilation error so I assumed it is true. Ok, let me ask you. What you will do if you see this in the gate. Will you not answer it?

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I hope gate people will not forget malloc() after typing a pointer declaration. It's like forgetting one right curly brace for a left curly brace.

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Mistake happens. Well hope so.

Answer 1

// Assuimg q pointer initialization :

    void find(struct Node *node) {
        struct Node *ptr,*q;
        
        q = (struct Node *)malloc(sizeof(struct Node));
        q->left  = NULL;
        q->right = NULL;

        if(node ==  NULL) return;
        
        find(nod->left);
        find(nod->right);
        
        1. ptr = node->left;
        2. q->value = node->value;
        3. node->left = q;
        4. node->left->left = ptr;
    }

for any non-null node:

Output :

 before 

  2
 1 3

 after 

        2
       / \
      /   \
     /     \
    2       3
   /       /  
  1       3    
 1              

Answer 2

Easy folow the procedure, code is doing -

Copying value of the node to q -> data.

saving the left node pointer to the ptr.

step 1 tree will be like

                           2

              1                     3

  (null) 1 ( not given)      

 

step 2

                           2

              1                     3

  (null) 1 ( not given)           (null) 3 ( not given) { 3 is the left child of 3 of earlier node}

 

step 3

 

                             2

                 2 ( not given)                           3

            1                             (null) 3 ( not given) { 3 is the left child of 3 of earlier node}

  (null) 1 ( not given)

Comments

I also thought the same. But for that they have to crater new node by using malloc and assign its adress, which I think made easy forgot to include.

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yes, I also think so . they were focused on tracing the program

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wrong QS ..are indeed wrong !