@Dave,
>From the definition of isomorphic trees(not in ques given), what i
know of is that one can be transformed into another. The above three
are then isomorphic to each other.
@Bugaboo, can you clarify what exactly do you mean by isomorphic
here?
On Aug 28, 9:25 pm, Dave <dave_and_da...@juno.com> wrote:
> @Naveet: So we have a question of semantics. Do these three trees have
> the same structure:
>
> a
> /
> b
> /
> c
>
> and
>
> a
> \
> b
> \
> c
>
> and
>
> a
> \
> b
> /
> c
>
> I say "no," but perhaps you say "yes."
>
> Dave
>
> On Aug 28, 9:35 am, Navneet <navneetn...@gmail.com> wrote:
>
>
>
>
>
>
>
> > Dave, that is why i have an OR condition between. Each side of OR has
> > two calls with AND in between.
>
> > Basically at any node, you will have to invoke with two combinations
> > ((left,left) AND (right,right) OR (left,right) AND (right,left))
>
> > Let me know if you think that's not required.
>
> > On Aug 28, 6:02 pm, Dave <dave_and_da...@juno.com> wrote:
>
> > > @Navneet: Don't we want both subtrees to be isomorphic?
>
> > > Dave
>
> > > On Aug 28, 6:40 am, Navneet <navneetn...@gmail.com> wrote:
>
> > > > Dave,
>
> > > > I think the last condition should be
>
> > > > return (AreIsomorphic(tree1->left, tree2->left) &&
> > > > AreIsomorphic(tree1->right,tree2->right)) ||
>
> > > > (AreIsomorphic(tree1->left, tree2->right) &&
> > > > AreIsomorphic(tree1->right,tree2->left))
>
> > > > On Aug 28, 3:39 pm, Ankur Garg <ankurga...@gmail.com> wrote:
>
> > > > > Daves solution looks cool to me...shud work :)
>
> > > > > Nice one Dave :)
>
> > > > > Regards
> > > > > Ankur
>
> > > > > On Sun, Aug 28, 2011 at 4:08 PM, Ankur Garg <ankurga...@gmail.com>
> > > > > wrote:
> > > > > > cant we just count the no of nodes in each level and compare them
> > > > > > with the
> > > > > > second one..
>
> > > > > > if the numbers are same trees can be said to be isomorphic
>
> > > > > > On Sun, Aug 28, 2011 at 3:54 AM, Dave <dave_and_da...@juno.com>
> > > > > > wrote:
>
> > > > > >> @Bugaboo: Use recursion. Assuming
>
> > > > > >> struct tree_node {
> > > > > >> tree_node *left;
> > > > > >> tree_node *right;
> > > > > >> int data;
> > > > > >> };
>
> > > > > >> int AreIsomorphic(tree_node tree1, tree_node tree2)
> > > > > >> {
> > > > > >> if( tree1 == NULL && tree2 == NULL )
> > > > > >> return TRUE; // both trees are null
> > > > > >> if( tree1 == NULL || tree2 == NULL)
> > > > > >> return FALSE; // one tree is null, the other is not
> > > > > >> return AreIsomorphic(tree1->left,tree2->left) &&
> > > > > >> AreIsomorphic(tree1->right,tree2->right);
> > > > > >> }
>
> > > > > >> Dave
>
> > > > > >> On Aug 27, 12:05 pm, bugaboo <bharath.sri...@gmail.com> wrote:
> > > > > >> > Considering the definition of binary tree isomorphism is the
> > > > > >> > following:
> > > > > >> > - 2 binary trees are isomorphic if they have the same structure
> > > > > >> > but
> > > > > >> > differ just by values.
>
> > > > > >> > What is the logic (or pseudo code) for checking if two binary
> > > > > >> > trees
> > > > > >> > are isomorphic?
>
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> > - Show quoted text -
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