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What is the time complexity of the optimal greedy postorder traversal solution for the Binary Tree Cameras problem, and why?

medium🪤 Complexity Trap Q13 of Q15
Tree: Depth-First Search - Binary Tree Cameras
What is the time complexity of the optimal greedy postorder traversal solution for the Binary Tree Cameras problem, and why?
AO(n) because each node is visited once in postorder traversal
BO(n^2) because each node's state depends on its children's states recursively
CO(n log n) due to recursive calls and state checks at each node
DO(h) where h is the height of the tree, since recursion depth equals height
Step-by-Step Solution

  1. Step 1: Analyze traversal visits

    The algorithm visits each node exactly once in a postorder manner, processing left and right children before the node itself.

  2. Step 2: Consider work per node

    Each node's processing is O(1) -- checking children's states and updating counters.

  3. Final Answer:

    Option A -> Option A

  4. Quick Check:

    Single DFS traversal over n nodes -> O(n) time [OK]
Quick Trick: Each node processed once with constant work [OK]
Common Mistakes:
MISTAKES
  • Assuming recursive calls multiply work to O(n^2)
  • Confusing recursion depth with total work
  • Thinking state checks add log n factor
Trap Explanation:
PITFALL
  • Recursion depth is O(h), but total nodes visited is n; confusing these leads to wrong complexity.
Interviewer Note:
CONTEXT
  • Tests understanding of DFS complexity and recursion effects.
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