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What is the time complexity of the BFS-based algorithm to compute the maximum depth of a binary tree with n nodes, and why might the following common misconception be incorrect? Options:

medium🪤 Complexity Trap Q13 of Q15
Tree: Depth-First Search - Maximum Depth of Binary Tree
What is the time complexity of the BFS-based algorithm to compute the maximum depth of a binary tree with n nodes, and why might the following common misconception be incorrect? Options:
AO(n), but with O(n) auxiliary space for the queue at the widest level
BO(n), because each node is visited exactly once in BFS
CO(n log n), because each level requires sorting nodes
DO(n^2), because each node is enqueued and dequeued multiple times
Step-by-Step Solution
Solution:

  1. Step 1: Identify time complexity

    BFS visits each node exactly once, so time complexity is O(n).

  2. Step 2: Identify space complexity and common misconception

    Queue can hold up to O(n) nodes at the widest level, so auxiliary space is O(n). The misconception is thinking nodes are processed multiple times, leading to O(n^2).

  3. Final Answer:

    Option A -> Option A

  4. Quick Check:

    Each node enqueued and dequeued once; max queue size O(n) [OK]
Quick Trick: BFS visits each node once; space depends on max level width [OK]
Common Mistakes:
MISTAKES
  • Assuming multiple visits per node
  • Confusing sorting with traversal
  • Ignoring queue space usage
Trap Explanation:
PITFALL
  • Option B looks correct but ignores space complexity; option C wrongly assumes repeated visits.
Interviewer Note:
CONTEXT
  • Checks understanding of BFS time and space complexity tradeoffs.
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