I recognized the underlying problem pretty fast since it's basically house robber on a tree.
Clarify the input format and constraints, then explain a tree DP approach where each node returns two values: max sum if the node is selected vs. not selected. Recurrence: if selected, add node value plus children's not-selected sums; if not selected, add max of children's selected/not-selected sums. Finally, return the max of the root's two states.
Pro tip: Mention that the 2D array representation is just a serialization; you can build the tree in O(n) using index arithmetic (children at 2i+1 and 2i+2) and then run the DP. This shows you separate parsing from the core algorithm.
Ask about the 2D array structure (e.g., rows/columns, placeholder values), node value ranges, and tree size. Confirm whether the tree is binary and if the array is level-order with nulls.
For each node, define two states: dp[node][0] = max sum in subtree when node is not selected; dp[node][1] = max sum when node is selected. Recurrence: dp[node][0] = sum(max(dp[child][0], dp[child][1])); dp[node][1] = node.val + sum(dp[child][0]).
Discuss trade-offs: recursion is simpler but may hit stack limits for deep trees; iterative post-order traversal avoids recursion but is more complex. Mention that for a complete binary tree, the array can be processed bottom-up without explicit tree construction.
Address empty tree, single node, negative values, and missing children. Analyze time and space complexity: O(n) time, O(n) space for DP table or O(h) for recursion stack.
Walk through a small example to verify recurrence. Mention space optimization: only need to keep two values per node, and for array representation, can compute in-place if allowed.
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