@Arun : yup ...rite...
so i guess this will work..once pointer has been updated for parent node.
there is no need of other point from children[i+1] to n. so adding one
more loop at the end.
for(i=0;i<n;i++)
{
pre=reverse(root->children[i],NULL);
if(pre)
{
pre->children[0]=root;
}
}
for(i=1;i<n;i++)
root->children[i] = NULL;
On Thu, Dec 22, 2011 at 1:45 PM, Arun Vishwanathan
<aaron.nar...@gmail.com>wrote:
> @atul: can you explain what this is doing?
> for(i=0;i<n;i++)
> {
>
> pre=reverse(root->children[i],NULL);
>
> if(pre)
> {
> pre->children[i]=root;
> }
> }
>
> when u do tree reversal I see that child points to parent( basically
> direction reversed). Now if
> 1
> / | \
> 2 3 4
> / / \
> 5 6 7
> / \
> 8 9
>
> in the last iteration along leftmost path , pointer to 8 is returned(pre)
> whose child is then assigned as 5 using pre->children[0]=root
> if 5 had another child say 'x' then after x is returned from thr recursive
> call, u wud assign pre->children[1]=root where root is pointer to 5
> here..but 8 and x have only one child which is 5 and indexes used for each
> are 0 and 1 instead of just 0..am i understanding it wrongly?
>
>
> On Wed, Dec 21, 2011 at 8:37 PM, atul anand <atul.87fri...@gmail.com>wrote:
>
>> adding one more check :- this one is updated
>>
>>
>> node *Reverse(node *root,node *pre)
>> {
>> flag=0;
>>
>> for i=0 to n
>> if (root->children[i]!=NULL)
>> {
>> flag=1;
>> break;
>> }
>> end for
>>
>> if( ! flag)
>> {
>> add root to the list;
>> return root;
>> }
>>
>> for(i=0;i<n;i++)
>> {
>> if(root->children[i]!=NULL)
>> {
>> pre=reverse(root->children[i],NULL);
>>
>> if(pre)
>> {
>> pre->children[i]=root;
>> }
>> }
>> }
>>
>> return root;
>>
>> }
>>
>>
>> On Thu, Dec 22, 2011 at 9:35 AM, atul anand <atul.87fri...@gmail.com>wrote:
>>
>>> adding break to first loop...just to avoid unnecessary iterations.
>>> if (root->children[i]!=NULL)
>>> {
>>> flag=1;
>>> break;
>>>
>>> }
>>> end for
>>>
>>>
>>> On Wed, Dec 21, 2011 at 10:59 PM, atul anand <atul.87fri...@gmail.com>wrote:
>>>
>>>> @shashank:
>>>>
>>>> yeah here is the algo , please me know if you find any bug:-
>>>>
>>>>
>>>> node *Reverse(node *root,node *pre)
>>>> {
>>>> flag=0;
>>>>
>>>> for i=0 to n
>>>> if (root->children[i]!=NULL)
>>>> {
>>>> flag=1;
>>>> }
>>>> end for
>>>>
>>>> if( ! flag)
>>>> {
>>>> add root to the list;
>>>> return root;
>>>> }
>>>>
>>>> for(i=0;i<n;i++)
>>>> {
>>>>
>>>> pre=reverse(root->children[i],NULL);
>>>>
>>>> if(pre)
>>>> {
>>>> pre->children[i]=root;
>>>> }
>>>> }
>>>>
>>>> return root;
>>>>
>>>> }
>>>>
>>>>
>>>>
>>>>
>>>> On Wed, Dec 21, 2011 at 1:08 PM, WgpShashank <
>>>> shashank7andr...@gmail.com> wrote:
>>>>
>>>>> @atul,, yeah , but can you write some proper psuedo code/ Algorithm
>>>>> then we can discus more about test cases.
>>>>>
>>>>> Thanks
>>>>> Shashank
>>>>>
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>
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