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.