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Given a binary tree where each node holds a non-negative integer representing money in a house, and you cannot rob two houses connected directly by an edge, which technique efficiently computes the maximum amount that can be robbed?

easy💻 Programming Q1 of Q15
Tree: Depth-First Search - House Robber III (On Tree)
Given a binary tree where each node holds a non-negative integer representing money in a house, and you cannot rob two houses connected directly by an edge, which technique efficiently computes the maximum amount that can be robbed?
ADynamic Programming on Trees using DFS with state tracking
BBreadth-First Search with level order traversal
CGreedy approach selecting nodes with maximum values first
DSorting nodes by value and picking top nodes ignoring edges
Step-by-Step Solution
Solution:

  1. Step 1: Understand the adjacency constraint

    You cannot rob two directly connected houses, so adjacent nodes cannot both be chosen.

  2. Step 2: Use DFS with two states per node

    For each node, compute two values: max if robbing this node, and max if not robbing it.

  3. Step 3: Combine results bottom-up

    Use recursion to combine children's states to decide the best choice at each node.

  4. Final Answer:

    Option A -> Option A

  5. Quick Check:

    Adjacency constraint handled correctly? Yes. [OK]
Quick Trick: Use DFS with rob/not_rob states for tree DP [OK]
Common Mistakes:
MISTAKES
  • Applying greedy approach ignoring adjacency
  • Using BFS without state tracking
  • Sorting nodes by value without considering edges
Trap Explanation:
PITFALL
  • Greedy or BFS approaches ignore adjacency constraints and fail on some trees.
Interviewer Note:
CONTEXT
  • Tests understanding of tree DP and adjacency constraints in House Robber III.
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