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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