@Sid +100
On Wed, Mar 21, 2012 at 10:20 AM, siddharam suresh
<siddharam....@gmail.com>wrote:
> 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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