@Nishaanth T1 has millions of nodes. Suppose all the nodes of T1 are equal to root of T2. Then u will have to check every where in T1. Putting height as constraint, u will have to check only those nodes whose height is equal to T2. It will reduce time complexity.
Well m not able to think of better time complexity, another way would be: Find Height of T2 ... O(k) //k is no. of nodes in T2 Find Height of each node in T1... O(N) //N is no. of nodes in T1 now if p nodes in T1 have height same as T2, then we can find if a subtree rooted at any of those p nodes are identical to T2 in O(pk) time. Thus total time complexity: O(N) + O(k) + O(pk). correct me if I am wrong.. -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
