seems correct with pre-order traversal.... if not give some example
On Wed, Mar 21, 2012 at 10:32 AM, Mahesh Thakur <tmahesh...@gmail.com>wrote:
> First Tree:
> 1
> 2 3
> 4 5 6 7
>
> Inorder traverse will be : 4251637
>
> Second Tree:
> 6
> 1 3
>
> Inorder traversal is 163.
>
> But they second tree is not subset. let me know if i got the question
> wrong.
>
> On Wed, Mar 21, 2012 at 10:27 AM, shady <sinv...@gmail.com> wrote:
>
>> @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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