@siddharam:

Performing inorder and pre/postorder will still have false positives.

Consider 2 trees one with root node "a" with node "b" as left child,
another tree with root node "a" with node "b" as its right child.

Anyway, for binary trees, I am aware of a recursive solution to find
out similarity or isomorphism. Not sure how this extends for a graph.
Always thought a DFS and/or BFS should suffice but apparently not.

On Sep 14, 1:43 am, siddharam suresh <[email protected]> wrote:
> bharath.sriram,
> perform inorder traversal and peorder/postorder traversal on both tree then
> compare both the result of two tree.
> Thank you,
> Sid.
>
>
>
>
>
>
>
> On Wed, Sep 14, 2011 at 10:22 AM, bugaboo <[email protected]> wrote:
> > This brings up another interesting question. How do you find out if 2
> > graphs are identical? (By identical, I mean exact similarity and NOT
> > isomorphism). Clearly, checking to see if both the DFS traversal and
> > BFS traversal match seem to have false positives as Bharathkumar
> > mentioned.
>
> > On Sep 13, 12:50 am, bharatkumar bagana <[email protected]>
> > wrote:
> > > @nishaanth:
> > > ex:
> > >    1
> > > 2    3
> > >          4
> > >             5
> > > bfs:12345
> > > dfs:12345
> > > branching factor of this tree  is not 1 ..........
>
> > > On Tue, Sep 13, 2011 at 9:38 AM, nishaanth <[email protected]>
> > wrote:
> > > > yes branching factor should be 1. it can be not equal to 1 only for the
> > > > penultimate node. by penultimate node i mean whose children are the
> > leaves
> > > > of the tree. rest all cases it should be 1.
>
> > > > On Tue, Sep 13, 2011 at 9:24 AM, siddharam suresh <
> > [email protected]
> > > > > wrote:
>
> > > >> cant say if there more than one leaf element
> > > >> still both the algo give same result
> > > >> Thank you,
> > > >> Sid.
>
> > > >> On Tue, Sep 13, 2011 at 7:52 AM, Sundi <[email protected]> wrote:
>
> > > >>> if the dfs and bfs of a graph is same, does it mean that if the
> > > >>> branching factor of a graph is one?
>
> > > >>> a>b>c>d
>
> > > >>> example: both dfs abd bfs are same here....
>
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