How do you find the maximum depth of a binary tree?

Question presented to candidate: "Given the root of a binary tree, write a Python function that returns its maximum depth: the number of nodes on the longest path from the root down to a leaf. Start with whatever is simplest, then tell me what happens if the tree is a very long chain."

What a strong answer should cover:

• The recurrence depth(None) = 0, depth(node) = 1 + max(depth(left), depth(right)), and that it is a postorder traversal.
• O(n) time because every node is visited once; O(h) space for the call stack, which is O(n) for a chain.
• An iterative alternative: BFS counting levels with collections.deque, or a stack of (node, depth) tuples.
• Clarifying nodes versus edges and the empty-tree answer; awareness of the minimum-depth trap with one-child nodes.
• Python specifics: the default recursion limit of 1000 means a 1000-deep chain raises RecursionError; use deque.popleft() not list.pop(0); tuple unpacking node, d = stack.pop() keeps the stack version short.

Clarifying questions expected:

• "Is depth counted in nodes or edges, and what should an empty tree return?"
• "Could the tree be very deep, like a linked list?"

Code / implementation expected: Yes — a recursive Python function, ideally followed by an iterative BFS or stack version.