Sorry for typo mistake in prev solution
2 solutions
1.
node* arr[100];
int j=0;
inorder(node * ptr)
{
if(ptr)
{
inorder(ptr->left);
arr[j++]=ptr;
inorder(ptr->right);
}
}
u will have all the addresses in inorder fashion.... now its easy to watch
any successor ... :P :P
2. best solution
//considering that there is 1 more field in the structure
typedef struct bin
{
struct bin* left;
int data;
struct bin* right;
struct bin* succ;
} node;
inorder(node* ptr)
{
if(ptr)
{
static int p=0;
inorder(ptr->left)
if(p) ptr->succ=prev; //here we are skipping 1st left most leaf
because it has no successor
p=1;
prev=ptr;
inorder(ptr->right);
}
}
simplest and short code :) :) :)
anyone have better code???
On Sat, Aug 6, 2011 at 8:24 PM, sagar pareek <[email protected]> wrote:
> 2 solutions
>
> 1.
>
> node* arr[100];
> int j=0;
>
> inorder(node * ptr)
> {
> if(ptr)
> {
> inorder(ptr->left);
> arr[j++]=ptr;
> inorder(ptr->right);
> }
> }
>
> u will have all the addresses in inorder fashion.... now its easy to watch
> any successor ... :P :P
>
> 2. best solution
> //considering that there is 1 more field in the structure
>
>
> typedef struct bin
> {
> struct bin* left;
> int data;
> struct bin* right;
> struct bin* succ;
> }
>
>
> inorder
> {
> if(ptr)
> {
> static int p=0;
> inorder(ptr->left)
> if(p) ptr->succ=prev; //here we are skipping 1st left most leaf
> because it has no successor
> p=1;
> prev=ptr;
> inorder(ptr->right);
> }
> }
>
>
> simplest and short code :) :) :)
>
> anyone have better code???
>
>
>
>
>
> On Sat, Aug 6, 2011 at 6:52 PM, UTKARSH SRIVASTAV <[email protected]
> > wrote:
>
>> sorry two cases only
>>
>>
>> On Sat, Aug 6, 2011 at 6:21 AM, UTKARSH SRIVASTAV <
>> [email protected]> wrote:
>>
>>> pseudo code
>>>
>>> three cases are possible
>>> 1.node has left and right child
>>> then inorder succesor will be leftmost child of right child
>>> 2. node has left child and no right child or no left and right chid
>>> if node is left child of it's parent then inorder succesor is it's
>>> parent only
>>> if node is right child of it's parent then keep on moving upwards
>>> until you find a parent which is left child of it's parent
>>> then it will be the inorder succesor....if you reach node then no
>>> inorder succesor
>>>
>>> --
>>> *UTKARSH SRIVASTAV
>>> CSE-3
>>> B-Tech 3rd Year
>>> @MNNIT ALLAHABAD*
>>>
>>>
>>
>>
>> --
>> *UTKARSH SRIVASTAV
>> CSE-3
>> B-Tech 3rd Year
>> @MNNIT ALLAHABAD*
>>
>> --
>> You received this message because you are subscribed to the Google Groups
>> "Algorithm Geeks" group.
>> To post to this group, send email to [email protected].
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>>
>
>
>
> --
> **Regards
> SAGAR PAREEK
> COMPUTER SCIENCE AND ENGINEERING
> NIT ALLAHABAD
>
>
--
**Regards
SAGAR PAREEK
COMPUTER SCIENCE AND ENGINEERING
NIT ALLAHABAD
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