This seems like a pretty straightforward problem: we essentially want to just count the maximum number of levels that the tree has. There's several ways to do this, but we'll need to traverse the tree.

Given that it's a binary tree, this means that there's up to two paths to move -- left and right. This fact simplifies things a bit, as traversal just requires navigating two children at each node.

During traversal, we can detect a leaf either when there is no left or right:

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if (!node.left && !node.right) {

  return;

}