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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