get the inorder traversal both tree (into strings) check weather one string
substring of other if yes then one tree is sub tree of other.
Thank you,
Sid.
phone:+91-8971070727,
+91-9916809982
On Wed, Mar 21, 2012 at 9:23 AM, HARSHIT PAHUJA <hpahuja.mn...@gmail.com>wrote:
> bool isSubtree(Tree * A,Tree *B)
> {
>
> if(!B) return true;
> if(!A)return false;
> if(A->data==B->data)
> return (isSubtree(A->left,B->left)
> && isSubtree(A->right,B->right));
> else
> return (isSubtree(A->left,B) && isSubtree(A->right,B));
> }
>
>
>
>
>
>
>
>
>
>
> }
>
> On Wed, Mar 21, 2012 at 2:33 AM, Don <dondod...@gmail.com> wrote:
>
>> bool equals(node *t1, node *t2)
>> {
>> return (t1 && t2) ? (t1->value == t2->value) && equals(t1->left, t2-
>> >left) && equals(t1->right, t2->right) : !t1 && !t2;
>> }
>>
>> bool check(node *t1, node *subtree)
>> {
>> return t1 ? equals(t1, subtree) || check(t1->left, subtree) ||
>> check(t1->right, subtree) : !subtree;
>> }
>>
>> On average this is the same as a traversal, but worst case could be
>> very slow. Imagine a large tree with millions of nodes, where all the
>> nodes = 1, and a somewhat smaller subtree with 100,000 nodes=1 and one
>> node at the far right of the tree = 2. It would require a lengthy
>> comparision at each node which would ultimately find no matching sub
>> tree.
>>
>> If they are binary search trees, it could be more efficient. Did you
>> mean to ask about binary search trees?
>>
>> Don
>>
>> On Mar 20, 7:24 am, Dheeraj Sharma <dheerajsharma1...@gmail.com>
>> wrote:
>> > How to check if one binary tree is a sub tree of other?
>> > Any Solution other then bruteforce?
>> > Prototype
>> > bool check(node *t1,node *subtree)
>> >
>> > --
>> > Sent from my mobile device
>> >
>> > *Dheeraj Sharma*
>>
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>>
>
>
> --
> HARSHIT PAHUJA
> M.N.N.I.T.
> ALLAHABAD
>
>
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