Two inorders would achieve the same thing without using an array. One pointer running inorder with LDR and other pointer running inorder with RDL. Compare the sum at the two nodes and then adjust them accordingly.
On Fri, Aug 6, 2010 at 2:11 PM, Manjunath Manohar <[email protected]>wrote: > the solution elegant..but is there any on the fly method by just exploiting > the BST property....by using left and right pointers > > -- > 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]<algogeeks%[email protected]> > . > 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 [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.
