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 <[email protected]> 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 <[email protected]> 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 <[email protected]> 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 <[email protected]> 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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