As per @Algoose's explanation, this can be found using the algorithm
to comapre if 2 binary trees are equal (quite often asked and found in
net). In this we generally go recursive and say
T1 is equal to T2
if T1=T2
and same holds for T1-Left, T2-Left (recursive call on left tree)
and same holds for T1-Right, T2-Right (recursive call on right tree).
We can use the same and change the calls as
if T1=T2
and same holds for T1-Left, T2-Right (recursive call on left tree)
and same holds for T1-Right, T2-Left (recursive call on right tree).
Sain
On Jun 9, 10:45 pm, saurabh gupta <sgup...@gmail.com> wrote:
> is-isomorphic(t1, t2) {
> t1ld = t1->left->data
> t2ld = t2->left->data
> //.....
>
> //base case for null or replace by sentinels and check
> if( t1ld == t2ld && t1rd == t2rd)
> return is-isomorphic(t1ld, t2ld) && return is-isomorphic(t1rd, t2rd)
> if (t1ld == t2rd && t1rd == t2ld)
> return is-isomorphic(t1ld, t2rd) && return is-isomorphic(t1rd, t2ld)
> return false;
>
>
>
>
>
> }
> On Wed, Jun 9, 2010 at 8:29 PM, divya jain <sweetdivya....@gmail.com> wrote:
> > @vadivel and anand
>
> > { 1,2,3 } is *isomorphic* to { 1,3,2 }
> > { 1,2,3,4,5,6,7 } is *isomorphic* to { 1,3,2,7,6,4,5 }
> > { 1,2,3,4,5,6,7 } is NOT *isomorphic* to { 1,3,2,4,5,6,7 }
>
> > so its nt necessary that right and left will interchange. it may or may
> > not. if all right and left are interchanged then it is mirror tree. i think
> > ur code is for mirror tree and not isomorphic tree..
>
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