can be done using some modification in postorder traversal call to left subtree will return that if left subtree is a BST of not call to right subtree will return that if right subtree is a BST of not if both subtrees are BST's check for curr and return its status
2 additional pass by ref. parameters will be needed for count and root of current best subtree On Sun, Jul 10, 2011 at 11:08 PM, Decipher <ankurseth...@gmail.com> wrote: > no of nodes > > -- > You received this message because you are subscribed to the Google Groups > "Algorithm Geeks" group. > To view this discussion on the web visit > https://groups.google.com/d/msg/algogeeks/-/MuNISw2QkYIJ. > > To post to this group, send email to algogeeks@googlegroups.com. > To unsubscribe from this group, send email to > algogeeks+unsubscr...@googlegroups.com. > For more options, visit this group at > http://groups.google.com/group/algogeeks?hl=en. > -- Sunny Aggrawal B-Tech IV year,CSI Indian Institute Of Technology,Roorkee -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
