The only way remains is to use the iterative method of traversing a BST in in-order (you have to use a Stack to keep track of father). The place where you print the value of the node, put a condition before that
if (!(--k)){ print value_of_node; } in the outermost loop where you check "while(no more nodes to traverse)" , add the condition that "( && k != 0)" ^^ This will keep a check on value of k after each loop run. Hope I am clear in explaining. -- 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/-/aHwinvhUrs4J. 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.