CODEWITHSUNDEEP
data-structures
Kumar Vishal
Kumar Vishal

Reputation: 121

insert method of binary tree

what's the difference in
[ insert_node(&(tmp->left),value);] VS [ tmp=tmp->right; insert_node(&(tmp),value);]

void insert_node(struct btree **bt,int value)
{
     btree *tmp= *bt, *r ;

     if(*bt==NULL)// first node 
     {
          (*bt)=(struct btree *)malloc(sizeof(btree));
          (*bt)->data=value;
          (*bt)->left=NULL;
          (*bt)->right=NULL;           
     }       
     else
     {
         if(value > tmp->data)
            insert_node(&(tmp->right),value);   
         else
            insert_node(&(tmp->left),value);                 
     }

      #if 0  //start
           OR  /** COMMENT START

         earlier I had below piece of code but not working 
         can any please explain what's the difference in  
         insert_node(&(tmp->left),value); VS [ tmp=tmp->right; insert_node(&(tmp),value);]
         COMMENT END **/ 

      else
     {
         if(value > tmp->data)
            tmp=tmp->right;  
         else
            tmp=tmp->left ;
            insert_node(&tmp,value);                 
     }
   #endif //end

}

Upvotes: 0

Views: 439

Answers (2)

Francis Upton IV
Francis Upton IV

Reputation: 19443

In the non-working case, you are giving it the address of your tmp variable, and therefore updating that pointer, which is on your stack and not going to be used for anything.

In the working case, you are giving the address of the actual pointer in your bt node which is what you want.

Upvotes: 1

IamIC
IamIC

Reputation: 18239

There is no semantic difference. #2 is storing the next node in a local variable, which will be a register, whereas #1 is storing the node into the register, but it won't be available as a local variable.

If you need to access the value again, use #2.

Upvotes: 0

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