@Dave : algo seems fine...but it seems to me that it would difficult to maintain both left to right and right to left parallel way :( :( . it would be gr8 if you can provided little bit of coded algorithm for it.
On Mon, Sep 3, 2012 at 10:36 AM, Dave <dave_and_da...@juno.com> wrote: > @Atul007: No need to destroy the BST. Perform two simultaneous inorder > traversals of the BST, one from left to right (the usual direction) and one > from right to left. At any stage you have selected two nodes. If the sum is > less than the given number, advance the left-to-right traversal; If the sum > is greater, advance the right-to-left traversal. Quit with success when the > sum equals the given number or with failure when the two traversals have > reached the same node. > Dave > > On Sunday, September 2, 2012 11:00:42 PM UTC-5, atul007 wrote: > >> convert BST to sorted DLL.. >> now it is a double linked list , so we can find 2 number in O(n) time by >> keeping 2 pointers(one at start and one at end) from sorted DLL. >> now convert DLL to BST. >> >> On Mon, Sep 3, 2012 at 1:32 AM, Navin Kumar <algorit...@gmail.com> wrote: >> MICROSOFT:Given a BST and a number. Find two node in a BST whose sum is >> equal to given number in O(n) time and O(1) space. >> >> -- > 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/-/oizd-5CSfuoJ. > > 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. > -- 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.
