Tree: Depth-First Search - House Robber III (On Tree)
What is the overall time complexity of the optimal DFS solution that returns two values (rob and not_rob) per node in the House Robber III problem on a binary tree with n nodes?
What is the overall time complexity of the optimal DFS solution that returns two values (rob and not_rob) per node in the House Robber III problem on a binary tree with n nodes?
medium📊 Complexity Q5 of Q15
Step-by-Step Solution
Solution:
Step 1: Analyze DFS traversal
The algorithm visits each node exactly once.Step 2: Constant work per node
At each node, it computes two values based on children's results, which is O(1) work.Step 3: Combine results
Since no repeated visits or nested traversals occur, total work is proportional to number of nodes.Final Answer:
Option B -> Option BQuick Check:
Each node processed once? Yes. [OK]
Quick Trick: DFS visits each node once with constant work [OK]
Common Mistakes:
MISTAKES- Assuming repeated visits cause O(n^2)
- Confusing with memoization overhead
- Thinking tree height affects complexity
Trap Explanation:
PITFALL- Mistaking repeated state computations or ignoring single-pass DFS.
Interviewer Note:
CONTEXT- Tests understanding of time complexity in tree DP solutions.
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