Dude, yeah I got the algo, but can u write java code for that

On Fri, Jul 30, 2010 at 6:03 PM, karthik ramanathan <[email protected]
> wrote:

> Do a BFS, by having a queue and keep inserting the nodes into it.
> You should know how many nodes are there in each level, for which you can
> have a variable to count the number of nodes in each level.
> The when you remove from your queue do connect these nodes till your count.
> You may need to use one more temp variable to not to lose the previous level
> node count when you compute the next level node count.
> Repeat the same for all the level.
>
> RK
>
>
>
> On Fri, Jul 30, 2010 at 11:21 AM, Amit Agarwal <[email protected]> wrote:
>
>> A simple queue implementation will do.
>>
>> -Regards
>> Amit Agarwal
>> blog.amitagrwal.com
>>
>>
>>
>>
>> On Fri, Jul 30, 2010 at 9:22 AM, Priyanka Chatterjee <[email protected]
>> > wrote:
>>
>>>
>>>
>>>   On 30 July 2010 02:59, Priyanka Chatterjee <[email protected]>wrote:
>>>
>>>>   Algo: 1. find height of tree
>>>>          2. do level order traversal
>>>>             i> at each level store the address of each tree node in the
>>>>  data part of a linked node and form linked list of the nodes
>>>>              ii> store the header of a linked list at a certain level in
>>>> an array
>>>>          3. return array.// gives final structure
>>>>
>>>>
>>>> complexity - T(n) =O(n)
>>>>                   S(n)=O(h+n ) //h=height of tree
>>>>
>>>> //CODE
>>>>
>>>> //tree structure
>>>> struct node {
>>>> int data;
>>>> struct node * left, *right};
>>>>
>>>> // linked list structure
>>>> struct linkNode{
>>>> struct node * data;
>>>> struct linkNode * next;
>>>> }
>>>>
>>>> struct linkNode** func(struct node * root){
>>>>
>>>> struct linkNode ** array;
>>>>
>>>> int i, h=height(root);
>>>> for(i=1;i<=h;i++)
>>>> array[i-1]=levelOrderTraversal(root, i);
>>>>
>>>> return array;// final tree structure
>>>> }
>>>>
>>>> //max height of tree
>>>>  int height(struct node *root){
>>>>  int hL=height(root->left);
>>>> int hR=height(root->right);
>>>> return 1+ HR>HL?HR:HL;
>>>> }
>>>>
>>>>
>>>>
>>>> struct nodelink* levelOrderTraversal(struct node*root, int level){
>>>> if(root==NULL) return NULL;
>>>>
>>>> if (level==1)
>>>>   return createLinkNode(root); // create a node of a singly l
>>>>
>>>>
>>>>   struct LinkNode *ptr;
>>>> if(level>1){
>>>> struct nodeLink * ptr1, *ptr2;
>>>> ptr1=levelOrderTraversal(root->left,level-1);
>>>> ptr2=levelOrderTraversal(root->right,level-1);
>>>>
>>>
>>> if(ptr1==NULL && ptr2==NULL) return NULL;
>>> if(ptr1==NULL) return ptr2;
>>> if(ptr2==NULL) return ptr1;
>>> ptr1->next=ptr2;
>>>
>>> return ptr2;
>>>
>>>> }
>>>>
>>>> }
>>>>
>>>>
>>>
>>>>  struct linkNode * createLinkNode(struct node * root){
>>>>
>>>> struct linkNode* newNode=(struct linkNode*) malloc(sizeof(struct
>>>> linkNode));
>>>>
>>>> newNode->data=root;
>>>>
>>>> newNode->next=NULL;
>>>>
>>>> }
>>>>
>>>>
>>>>
>>>> --
>>>> Thanks & Regards,
>>>> Priyanka Chatterjee
>>>> Final Year Undergraduate Student,
>>>> Computer Science & Engineering,
>>>> National Institute Of Technology,Durgapur
>>>> India
>>>> http://priyanka-nit.blogspot.com/
>>>>
>>>
>>>
>>>
>>> --
>>> Thanks & Regards,
>>> Priyanka Chatterjee
>>> Final Year Undergraduate Student,
>>> Computer Science & Engineering,
>>> National Institute Of Technology,Durgapur
>>> India
>>> http://priyanka-nit.blogspot.com/
>>>
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>>
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-- 
Regards,
vineel.

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