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12. Maximum Depth of Binary Tree

easyAsked at Dropbox

Compute the maximum depth of a binary tree; Dropbox uses it as a primer for bounding recursion depth on deeply nested folder trees in user accounts.

By Alex Chen, Founder, InterviewChamp.AI · Last verified

Problem

Given the root of a binary tree, return its maximum depth (the number of nodes along the longest path from root to leaf).

Constraints

  • 0 <= nodes <= 10^4
  • -100 <= Node.val <= 100

Examples

Example 1

Input
root=[3,9,20,null,null,15,7]
Output
3

Example 2

Input
root=[1,null,2]
Output
2

Approaches

1. BFS level count

Queue each level, count rounds.

Time
O(n)
Space
O(w)
if(!root) return 0;let q=[root],d=0;
while(q.length){d++;q=q.flatMap(n=>[n.left,n.right].filter(Boolean));}
return d;

Tradeoff:

2. DFS recursion

Depth(node) = 0 if null else 1 + max(depth(left), depth(right)). Cleanest one-liner pattern.

Time
O(n)
Space
O(h)
function maxDepth(root) {
  if (!root) return 0;
  return 1 + Math.max(maxDepth(root.left), maxDepth(root.right));
}

Tradeoff:

Dropbox-specific tips

Dropbox grades on whether you flag iterative-vs-recursive for skewed trees — they want you to mention that a 100k-deep folder tree would blow the call stack.

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