@ sanchit
m dong inorder traversal and at every step chking whether the node's p
pointer is pointing to its inorder predecesor, which is temp or not and
making it NULL otherwise. when count is 0 the node do not hv any predecessor
so m directly pointing that to NULL.
*please see the below alg0 , the abv algo by me is incomplete (sorry :(
) *
int count=0; //global
void modified_inorder(node *root){
if(root!=NULL){
modified_inorder(root->left);
node *temp;
if(count==0){
root->p=NULL;
temp=root;
count++;
}
else{
if(root->p!=temp){
root->p=NULL;
}
temp=root;
modified_inorder(root->right);
}
}
}
On Mon, Feb 14, 2011 at 9:45 PM, sanchit mittal <[email protected]> wrote:
> explain algo instead of writing the code.....
> thanx
>
>
> On Mon, Feb 14, 2011 at 9:28 PM, jalaj jaiswal
> <[email protected]>wrote:
>
>> @tushar that would modify the tree structure
>>
>> here is a different approach
>>
>> int count=0; //global
>> void modified_inorder(node *root){
>> if(root!=NULL){
>> modified_inorder(root->left);
>> node *temp;
>> if(count==0){
>> root->p=NULL;
>> temp=root;
>> count++;
>> }
>> else{
>> if(root->p!=temp){
>> root->p=NULL;
>> }
>> temp=root;
>>
>> }
>>
>> }
>>
>> On Mon, Feb 14, 2011 at 7:59 PM, Tushar Bindal <[email protected]>wrote:
>>
>>> I think the following algo should work:
>>> 1. Create a DLL of the inorder traversal of the tree
>>> 2. for each node, check whether P of that node points to the previous
>>> node in the DLL or not.
>>> 3. If not, assign it value NULL
>>>
>>>
>>>
>>> --
>>> Tushar Bindal
>>> Computer Engineering
>>> Delhi College of Engineering
>>> Mob: +919818442705
>>> E-Mail : [email protected], [email protected]
>>>
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>>
>>
>>
>> --
>> With Regards,
>> *Jalaj Jaiswal* (+919019947895)
>> Software developer, Cisco Systems
>> B.Tech IIIT ALLAHABAD
>>
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>
>
>
> --
> Sanchit Mittal
> Second Year Undergraduate
> Computer Engineering
> Delhi College of Engineering
> ph- +919582414494
>
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>
--
With Regards,
*Jalaj Jaiswal* (+919019947895)
Software developer, Cisco Systems
B.Tech IIIT ALLAHABAD
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