Hmm, weird, the format is wrong again, anyway it looks like this

void printLevel(Node* n, int level, bool order){

if(n){
if(level==1) “print the data here” […]

Format issue in the previous post, here it is

void printLevel(Node* n, int level, bool order){

if(n){
if(level==1) cout<data<left, level-1, order);
printLevel(n->right, level-1, […]

A solution without extra space (the two stacks) is to write a helper function which prints a specific level of tree in a pre-defined order:

if order is true, left to right, else right to left

void […]

I use a Dynamic Programming strategy and I think it is very clear.

I define f(n) as the largest number of nodes in a binary search tree rooted EXACTLY at n. Here as usual, this BST doesn’t need to contain all […]

The naive approach is obviously not correct, but the idea of using min and max to keep track of the boundary scope of a BST is brilliant.