Option 3 is correct :-
suppose we have tree
a
/ \
/ \
b c
/ \ / \
d e f g
now it will first delete g
then it will delete f
now why would it delete c ?
we are deleting f and g because both left and right are NULL of f and g
but still c->left and c->right has address of f and g respectively (dangling
pointer ) but if we check the condition (c->left==NULL&&c->right==NULL) , it
will not satisfy hence c will not be deleted...
so we we have to do c->left=NULL and c->right=NULL before deleting
pls correct me if i m wrong
On Mon, Aug 8, 2011 at 11:23 PM, *$* <[email protected]> wrote:
> I guess option 1 is correct
>
> Thx,
> --Gopi
>
>
> On Mon, Aug 8, 2011 at 11:07 PM, siddharam suresh <[email protected]
> > wrote:
>
>> func( Node *node){
>>
>> if( node->left == NULL && node->right == NULL )
>>
>> delete(node);
>>
>> if(node->right != NULL)
>>
>> func( node->right);
>>
>> if(node->left != NULL)
>>
>> func( node->left);
>>
>>
>> }
>>
>> for this code option 3 is correct
>>
>> Thank you,
>> Siddharam
>>
>>
>>
>> On Mon, Aug 8, 2011 at 11:03 PM, rajul jain <[email protected]>wrote:
>>
>>> This is what i am trying to say to debabrata
>>>
>>>
>>> On Mon, Aug 8, 2011 at 10:56 PM, aditi garg
>>> <[email protected]>wrote:
>>>
>>>> @debrata : bt remember its a recursive function...once the leaf node
>>>> gets deleted thn it will move bak to the node and thn to the left of it
>>>> fr eg
>>>> a
>>>> / \
>>>> b c
>>>> / \ / \
>>>> d e f g
>>>> frst g will be deleted...then f thn it wil delete c and so on...i think
>>>> the ans shud be 1...correct me if i am wrong...
>>>> On Mon, Aug 8, 2011 at 10:49 PM, Debabrata Das <
>>>> [email protected]> wrote:
>>>>
>>>>> check this condition
>>>>> if( node->left == NULL && node->right == NULL )
>>>>> it is true only for leaves node
>>>>>
>>>>> On Mon, Aug 8, 2011 at 10:46 PM, rajul jain <[email protected]>
>>>>> wrote:
>>>>> > I have also made same answer on first look but read 3rd option
>>>>> correctly it
>>>>> > say deletion of leaves from right to left not internal nodes.
>>>>> >
>>>>> > On Mon, Aug 8, 2011 at 10:36 PM, Debabrata Das
>>>>> > <[email protected]> wrote:
>>>>> >>
>>>>> >> i think 3
>>>>> >>
>>>>> >> On Mon, Aug 8, 2011 at 10:32 PM, rajul jain <[email protected]
>>>>> >
>>>>> >> wrote:
>>>>> >> > got it thanks
>>>>> >> >
>>>>> >> > On Mon, Aug 8, 2011 at 10:30 PM, Akash Mukherjee <
>>>>> [email protected]>
>>>>> >> > wrote:
>>>>> >> >>
>>>>> >> >> i think its 1 though .......
>>>>> >> >>
>>>>> >> >> On Mon, Aug 8, 2011 at 10:21 PM, rohit <[email protected]>
>>>>> wrote:
>>>>> >> >>>
>>>>> >> >>> What will the following code snippet do, when is it passed the
>>>>> root of
>>>>> >> >>> a binary tree ?
>>>>> >> >>> func( Node *node){
>>>>> >> >>>
>>>>> >> >>> if(node->right != NULL)
>>>>> >> >>>
>>>>> >> >>> func( node->right);
>>>>> >> >>>
>>>>> >> >>> if(node->left != NULL)
>>>>> >> >>>
>>>>> >> >>> func( node->left);
>>>>> >> >>>
>>>>> >> >>> if( node->left == NULL && node->right == NULL )
>>>>> >> >>>
>>>>> >> >>> delete(node);
>>>>> >> >>>
>>>>> >> >>> }
>>>>> >> >>>
>>>>> >> >>> Pick choice
>>>>> >> >>> Delete the tree from bottom to top
>>>>> >> >>>
>>>>> >> >>> Delete the tree from top to bottom
>>>>> >> >>>
>>>>> >> >>> Delete the leaf nodes from right to left
>>>>> >> >>>
>>>>> >> >>> Delete the leaf nodes from left to right
>>>>> >> >>>
>>>>> >> >>> I think it is 3
>>>>> >> >>>
>>>>> >> >>> --
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>>>>>
>>>>
>>>>
>>>> --
>>>> Aditi Garg
>>>> Undergraduate Student
>>>> Electronics & Communication Divison
>>>> NETAJI SUBHAS INSTITUTE OF TECHNOLOGY
>>>> Sector 3, Dwarka
>>>> New Delhi
>>>>
>>>>
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>
>
>
> --
> Thx,
> --Gopi
>
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--
**Regards
SAGAR PAREEK
COMPUTER SCIENCE AND ENGINEERING
NIT ALLAHABAD
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