and to count no. of paths, you can do this
void printPath(node *p,node *child)
{
static int iCountPath = 0; // or pass it as an argument non-const ref.
if( p && child)
{
++iCountPath;
std::cout << "Path: " << p->data << " => " << child->data <<
std::endl;
}
}
--
Amitesh
On Thu, Jun 28, 2012 at 11:01 AM, Amitesh Singh <singh.amit...@gmail.com>wrote:
> void printPath(node *p,node *child)
> {
> if( p && child)
> std::cout << "Path: " << p->data << " => " << child->data <<
> std::endl;
> }
>
> use this before you assign 'current' to its children.
> e.g.
>
> printPath(p,p->left)
> or
> printPath(p,p->right);
>
>
> --
> Amitesh
>
>
>
>
> On Mon, Jun 25, 2012 at 11:34 PM, Nishant Pandey <
> nishant.bits.me...@gmail.com> wrote:
>
>>
>> thiss is iterative inorder code i am using for printing all the paths of
>> the Btree .
>>
>> but i am not able to manipulate the lengths properly when its perform pop
>> up so that i
>> can print the paths properly ..
>>
>> Can any one help by modifying the changes in this code .
>>
>>
>>
>> void inorder(struct node *root)
>> {
>> stack<struct node*>stack_temp;
>>
>> struct node *current,*temp;
>> int path[20];
>> int i =0 ,j=0;
>> if(!root)return ;
>>
>> current=root;
>> bool flag=false;
>> //bool ctrl=false;
>> while(!flag)
>> {
>>
>> if(current)
>> {
>> //cout<<current->data<<endl;
>> stack_temp.push(current);
>> path[i++]=current->data;
>> if(current->left == NULL && current->right == NULL)
>> {
>> for(j=0;j<i;j++)
>> cout<<path[j]<<" ";
>> //i--;
>>
>> }
>>
>> cout<<endl;
>> current=current->left;
>>
>> }
>> else
>> {
>> if(stack_temp.empty())
>> {
>> flag=true;
>> }
>> else
>> {
>> current=stack_temp.top();
>> stack_temp.pop();
>>
>> i--;
>>
>>
>> //if(current->right)
>> //cout<<current->data<<endl;
>> current=current->right;
>> if(current)
>> i++;
>> }
>>
>> }
>>
>> }
>>
>>
>> }
>>
>>
>>
>> --
>> Cheers,
>>
>> Nishant Pandey |Specialist Tools Development |
>> npan...@google.com<gvib...@google.com> |
>> +91-9911258345
>>
>>
>> --
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>
>
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