HackerRank Binary search tree validation

Question

From here:

def checkBST(root): prev_val = -1 for node in in_order_sort(root): if node.data <= prev_val: return False prev_val = node.data return True def in_order_sort(node): if node.left: yield from in_order_sort(node.left) yield node if node.right: yield from in_order_sort(node.right) 

Looking for any suggestions on improving this. It's pretty concise.

The input data is constrained between \$0\$ and \$10^4\$ so I can "get away" with setting the initial value to -1. The input func name checkBST is also predefined.

It seems that short of knowing you can validate a binary tree via an in-order traversal this would get complicated, but knowing that makes it straightforward?

Answer 1

The prev_val handling is slightly clumsy. I would personally prefer using the pairwise() recipe from itertools. You could then replace the loop altogether with all(…), which more clearly expresses your intentions.

I would also prefer to see the generator yield node.data instead of yield node.

from itertools import tee def checkBST(root): a_iter, b_iter = tee(in_order_traversal(root)) next(b_iter, None) return all(a <= b for a, b in zip(a_iter, b_iter)) def in_order_traversal(node): if node.left: yield from in_order_sort(node.left) yield node.data if node.right: yield from in_order_sort(node.right) 

Answer 2

I'd leverage python's builtin functions by flattening the tree to a list, then checking if it's sorted in ascended order and whether it has any duplicates:

def checkBST(root): flat = flatten(root) return flat == sorted(flat) and len(flat) == len(set(flat)) def flatten(tree): if tree: return flatten(tree.left) + [tree.data] + flatten(tree.right) else: return [] 

This will almost certainly be slower though.