a
/ \
/ \
b c
/ \ / \
d e f g
for the given tree, the ans depends on the way one has implemented delete
function.... as correctly mentioned by sagar if it only deletes and doesnt
care of returning null to c->right after deleting g then the answer will be
3, and if it delete is properly implemented then it deletes as
g, f, c, e, d, b, a
On Tue, Aug 9, 2011 at 12:08 AM, aditi garg <aditi.garg.6...@gmail.com>wrote:
> when f and g have been deleted wudnt the last condition left==null&&
> right==null be satisfied fr c as well?
>
>
> On Tue, Aug 9, 2011 at 12:01 AM, sagar pareek <sagarpar...@gmail.com>wrote:
>
>> 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, *$* <gopi.komand...@gmail.com> wrote:
>>
>>> I guess option 1 is correct
>>>
>>> Thx,
>>> --Gopi
>>>
>>>
>>> On Mon, Aug 8, 2011 at 11:07 PM, siddharam suresh <
>>> siddharam....@gmail.com> 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 <rajuljain...@gmail.com>wrote:
>>>>
>>>>> This is what i am trying to say to debabrata
>>>>>
>>>>>
>>>>> On Mon, Aug 8, 2011 at 10:56 PM, aditi garg <aditi.garg.6...@gmail.com
>>>>> > 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 <
>>>>>> debabrata.barunhal...@gmail.com> 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 <rajuljain...@gmail.com>
>>>>>>> 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
>>>>>>> > <debabrata.barunhal...@gmail.com> wrote:
>>>>>>> >>
>>>>>>> >> i think 3
>>>>>>> >>
>>>>>>> >> On Mon, Aug 8, 2011 at 10:32 PM, rajul jain <
>>>>>>> rajuljain...@gmail.com>
>>>>>>> >> wrote:
>>>>>>> >> > got it thanks
>>>>>>> >> >
>>>>>>> >> > On Mon, Aug 8, 2011 at 10:30 PM, Akash Mukherjee <
>>>>>>> akash...@gmail.com>
>>>>>>> >> > wrote:
>>>>>>> >> >>
>>>>>>> >> >> i think its 1 though .......
>>>>>>> >> >>
>>>>>>> >> >> On Mon, Aug 8, 2011 at 10:21 PM, rohit <rajuljain...@gmail.com>
>>>>>>> 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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>>>>>
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>>>
>>>
>>>
>>> --
>>> Thx,
>>> --Gopi
>>>
>>>
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>>
>>
>>
>> --
>> **Regards
>> SAGAR PAREEK
>> COMPUTER SCIENCE AND ENGINEERING
>> NIT ALLAHABAD
>>
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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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