how about using the threaded binary tree?
On 6 August 2011 20:25, sagar pareek <[email protected]> wrote:
> 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
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>>>
>>
>>
>>
>> --
>> **Regards
>> SAGAR PAREEK
>> COMPUTER SCIENCE AND ENGINEERING
>> NIT ALLAHABAD
>>
>>
>
>
> --
> **Regards
> SAGAR PAREEK
> COMPUTER SCIENCE AND ENGINEERING
> NIT ALLAHABAD
>
> --
> You received this message because you are subscribed to the Google Groups
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