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Coding workflows · Strict binary-search-tree ordering
03
A concrete starting point
Validate integer search-tree nodes with two child references. Check empty trees and ancestor bounds, returning false for duplicate keys while rejecting cycles, shared children, malformed nodes or over 1000 nodes.
Define the change
Describe the input, expected output and boundaries that must stay intact.
Inspect the work
Review the proposed edits and any commands before relying on their result.
Check the outcome
Compare actual behavior with the expected result. If a check fails, investigate it before applying more changes.
Strict binary-search-tree ordering
A concrete starting point
def is_strict_bst(root):
pending, seen, checks = [(root, None, None)], set(), []
while pending:
node, low, high = pending.pop()
if node is None:
continue
if type(node) is not list or len(node) != 3 or type(node[0]) is not int:
raise ValueError("tree.node")
if id(node) in seen or len(seen) == 1000:
raise ValueError("tree.structure")
seen.add(id(node))
value, left, right = node
checks.append((value, low, high))
pending.extend([(right, value, high), (left, low, value)])
return all((low is None or low < value) and
(high is None or value < high) for value, low, high in checks)The example is a starting point for a real project, not a claim that generated code is already tested. Adjust paths, dependencies and expected behavior to match your application.