Is the depth counted in nodes or in edges?

Nodes: a single node has depth 1 and the empty tree 0, so [4, 2, 7, null, null, 6, 9, 5] gives 4. A definition in edges (often called the height) is one less for every non-empty tree and -1 for the empty one; return -1 instead of 0 in the base case to get it.

Can the recursive maxDepth throw RangeError?

Yes, on a tree deep enough. Every pending call keeps a frame on the engine's fixed-size call stack, and a tree that leans entirely one way makes the recursion as deep as the node count, so a chain of tens of thousands of nodes throws RangeError: Maximum call stack size exceeded. The level loop has no such limit. The cases here stay within 500 levels, so both versions pass.

Why build a new array per level instead of using a queue with shift()?

shift() is defined as moving every remaining element down one slot, so on a wide level it can cost O(n) per call. A fresh next array per row keeps the whole walk linear, and it also marks where one level ends without counting. A single array walked with a read index works too, as long as you remember the index where each level stops.

Why does my maxDepth return NaN?

Usually the base case returns nothing: if (root === null) return; hands back undefined, and Math.max(undefined, 0) is NaN, which then spreads up through every 1 + NaN. Return 0 explicitly. A related trap: if a missing child were undefined rather than null, root === null would let it through and reading root.left would throw TypeError; root == null matches both, though in this harness a missing child is always null.

How do I get the minimum depth instead?

The minimum depth is the shortest path to a leaf, and a node with one child is not a leaf. Replacing Math.max with Math.min is wrong: on [10, null, 20, null, 30] it returns 1 because the missing left child counts as a path of zero, while the answer is 3. With the level loop, return the depth of the first node that has no children.

Maximum Depth of a Binary Tree in JavaScript: Recursive and BFS Solutions

max_depth.js

JavaScript 26.10

Not run yet

+ 7 hidden

Input

root = [4, 2, 7, null, null, 6, 9, 5]

Expected

The longest path is 4 → 7 → 6 → 5, four nodes. The branch through 2 ends after two, and 9 after three.

1 · Recursive DFS: one more than the deeper subtree

2 · Iterative BFS: count the levels with a queue

1 · Recursive DFS: one more than the deeper subtree

2 · Iterative BFS: count the levels with a queue

1 · Recursive DFS: one more than the deeper subtree

2 · Iterative BFS: count the levels with a queue

1 · Recursive DFS: one more than the deeper subtree

2 · Iterative BFS: count the levels with a queue

1 · Recursive DFS: one more than the deeper subtree

2 · Iterative BFS: count the levels with a queue

1 · Recursive DFS: one more than the deeper subtree

2 · Iterative BFS: count the levels with a queue

1 · Recursive DFS: one more than the deeper subtree

2 · Iterative BFS: count the levels with a queue

1 · Recursive DFS: one more than the deeper subtree

2 · Iterative BFS: count the levels with a queue

1 · Recursive DFS: one more than the deeper subtree

2 · Iterative BFS: count the levels with a queue

1 · Recursive DFS: one more than the deeper subtree

2 · Iterative BFS: count the levels with a queue

1 · Recursive DFS: one more than the deeper subtree

2 · Iterative BFS: count the levels with a queue

1 · Recursive DFS: one more than the deeper subtree

2 · Iterative BFS: count the levels with a queue

1 · Recursive DFS: one more than the deeper subtree

2 · Iterative BFS: count the levels with a queue

1 · Recursive DFS: one more than the deeper subtree

2 · Iterative BFS: count the levels with a queue

1 · Recursive DFS: one more than the deeper subtree

2 · Iterative BFS: count the levels with a queue