Hi,
    But in your second while( ) loop when the check 'root->left = NULL'
executes 'root' is already NULL, because only then the first loop
terminates. So you need to insert the following statement inbetween the
loops

    root = pop( )

    Also your solution assumes that the value at the nodes are of 'single
character'. Say, if the values are of multiple character strings, then
instead of 'string_reverse( )' the function 'reverse_words_in_string( )'.

   Other than this the solution should be perfect. Correct me if I'm wrong.

Thanks and Regards,
K.V.Chandra Kumar.

On 07/09/2007, anshu <[EMAIL PROTECTED]> wrote:
>
>
> An alternative algorithm that works without the explored() function
> could be as under.
> Just written a rough algorithm, there are a few more optimizations in
> loop possible,..
> The basi idea used here is -
> Post order mean "Data-Left-Right"
> If pre-order algorithm be executed with the difference that instead of
> left we explore right.
> The final order obtained when reversed will give post order.
>
>
> post_order(root)
> {
>     string_reverse( func(root));
> }
>
>
> func(root)
> {
>    do {
>           while(root != NULL)
>           {
>                 print(root->data);
>                 push(root);
>                 root=root->right;
>           }
>           while( root->left=NULL || stack ! empty)
>                   root=pop()
>           if(root->left !=NULL)
>           {
>                 root=root->left
>           }
>
>    }while( stack ! empty);
>
> }
>
>
> On Aug 28, 8:39 am, "chandra kumar" <[EMAIL PROTECTED]>
> wrote:
> > Hi,
> >     Need more details about explored(  Node * ) function,
> >
> >     Consider the "NULL" input
> >
> >     if your explored( NULL ) returns "true" then I guess that every
> thing
> > works fine, and also most of your checks could be eliminated ( code will
> > become simpler )
> >     if your explored( NULL ) returns "false", then the same case as in
> my
> > previous mail will result in wrong answer.
> > *        if((s->left == null && s->right==null)
> > ||(explored(s->left)&&explored(s->right)) *
> > *        ---as s->left == null && explored( s->right )   ( and vice
> versa
> > are not there )*
> >
> > Correct me if I'm wrong.
> >
> > Thanks and Regards,
> > K.V.Chandra Kumar.
> >
> > On 28/08/07, MD <[EMAIL PROTECTED]> wrote:
> >
> >
> >
> > > I think first s=pop() in while is not the right approach. This is an
> > > alternate approach where explored() checks if the node is visited or
> > > not... hence discarding that path.. and I think the following handles
> > > the null conditions as well.. (ex given by chandra)
> >
> > > void postOrderTraversal(Tree *root)
> > > {
> > > node * previous = null;
> > > node * s = root;
> > > push(s);
> >
> > > while( stack is not empty)
> > > {
> > > if(s->left && !explored(s->left))  //explored check if the node was
> > > previously visited
> > >    {push(s->left);
> > >    s=s->left}
> > > else
> > >   {if(s->right && !explored(s->right))
> > >       {push(s->right);
> > >        s=s->right;}
> > >   }
> >
> > >   if((s->left == null && s->right==null) ||(explored(s-
> > > >left)&&explored(s->right)) //last level-child or both childern are
> > > explored
> > >   { s = pop(); //
> > >     print(s->data);
> > >     s= pop(); //POP Again....point s to next element.
> > >   }
> > > }//end of while
> >
> > > }
> >
> > > On Aug 24, 6:17 am, "Phani Kumar Ch. V." <[EMAIL PROTECTED]> wrote:
> > > > Hi all,
> >
> > > > Please let me know if this pseudo code gives correct solution for
> > > iterative
> > > > post-order traversal of a binary tree.
> > > > ----------------------------------------------------
> > > > void postOrderTraversal(Tree *root)
> > > > {
> > > > node * previous = null;
> > > > node * s = null;
> > > > push(root);
> > > > while( stack is not empty )
> > > > {
> > > >   s = pop();
> >
> > > >   if(s->right == null and s->left == null)
> > > >   {
> > > >     previous = s;
> > > >     process node s;
> > > >   }
> > > >   else
> > > >   {
> > > >     if( s->right == previous or s->left == previous )
> > > >     {
> > > >       previous = s;
> > > >       process node s;
> > > >     }
> > > >     else
> > > >     {
> > > >       push( s );
> > > >       if(s->right) { push(s->right); }
> > > >       if(s->left)  { push(s->left);  }
> > > >     }
> > > >   }}
> >
> > > > -----------------------
> > > > Regards
> > > > Phani
>
>
> >
>

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