Maximum Depth of Binary Tree

Given the root of a finite binary tree, return its maximum depth: the number of nodes on the longest path from the root to a leaf. Return 0 for an empty tree.

Function
maximumDepth(root: TreeNode) → int

Examples
Example 1
root = [3,9,20,null,null,15,7]
return = 3
The longest root-to-leaf path contains three nodes.
Example 2
root = []
return = 0
An empty tree has depth zero.

Constraints
The tree contains at most 100000 nodes.
Every node value fits in a signed 32-bit integer.
The input is a finite acyclic binary tree.

The trick is one line of recursion: depth(node) = 0 if null, otherwise 1 + max(depth(left), depth(right)). That's depth-first search, and it's the whole solution. The pitfall is the constraint. A tree with 100000 nodes can be a straight chain, and a recursive solution may blow the call stack in some languages. If that worries you, use an iterative BFS: push the root into a queue, process one level at a time, and count levels. Same O(n) time, and the queue holds at most one level so memory stays bounded. Also handle the empty tree first and return 0. Count nodes, not edges, so example 1 gives 3, not 2. If you blank on the live LinkedIn OA, StealthCoder runs invisibly and gives you the recursive or BFS version to check against.

Drill it cold or hedge it with StealthCoder. Either way, don't walk into the OA hoping you remember the trick.