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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