How about this
Stack<tree*> st;
CheckSubtree(tree *bigTree, tree *smallTree)
{
if(bigTree == NULL) return;
if(bigTree->data == smallTree->data) st.Push(bigTree);
CheckSubtree(bigTree->left, smallTree);
CheckSubtree(bigTree->right, smallTree)
}
bool compareSubtrees(tree* t1, tree* t2)
{
if(t1 == t2) return true;
if(t1->data != t2->data) return false;
return compareSubtrees(t1->left, t2->left)
&& compareSubtrees(t1->right, t2->right);
}
Vector<bool> checkSubtreeMain(tree* big, tree* small)
{
CheckSubtree(tree *bigTree, tree *smallTree);
tree *node;
vector<bool> retValues;
while((node = st.Pop()) != null)
{
bool ret = compareSubtrees(node, small);
if(ret) retValues.push(ret);
}
printf(" subtree present @ "+retValues.size());
return retValues;
}
On Wednesday, 24 October 2012 11:16:01 UTC+5:30, rahul sharma wrote:
>
> We could create a string representing the inorder and preorder traversals.
> If T2’s preorder traversal is a substring of T1’s preorder traversal, and
> T2’s inorder traversal is a substring of T1’s inorder traversal, then T2 is
> a substring of T1
>
>
> any other method??
> we can also do by folloowing
>
> 1. t1 null
> false
> 2. if t2 null
> treu
> 3. if data of both are equal call for identical fxn
>
> just rough idea
>
> anyother approach?
>
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