@shady consider this case
Main tree 1
subtree 1
2 3
2
4
3
preorder of Main= 1234 subtree= 123 but not subtree
On Wed, Mar 21, 2012 at 1:24 PM, shady <sinv...@gmail.com> wrote:
> 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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