My first instinct was to treat it like a graph coloring problem and I wasted probably three minutes going down that road.
Recognize this as the 'Maximum Weight Independent Set on a Tree' problem and solve it with dynamic programming. For each node, compute two values: the maximum sum when the node is included (then children must be excluded) and when it is excluded (children can be either included or excluded). Use post-order traversal to combine these values bottom-up, returning the maximum of the two values at the root.
Pro tip: Clarify edge cases upfront (empty tree, single node, negative values) and mention that the DP can be implemented iteratively to avoid recursion depth issues. Also, briefly discuss how the solution would change if the tree were a general graph (NP-hard), showing awareness of problem constraints.
Confirm that the tree is binary, values are non-negative, and we need the maximum sum with no parent-child selections. Ask about input size to determine if recursion depth is a concern.
For each node, define two states: include[node] = max sum in subtree when node is selected; exclude[node] = max sum when node is not selected. The answer for the subtree is max(include, exclude).
If node is included, its children must be excluded: include[node] = node.val + sum(exclude[child]). If node is excluded, children can be either included or excluded: exclude[node] = sum(max(include[child], exclude[child])).
Use post-order DFS (recursive or iterative) to compute states bottom-up. Return max(include[root], exclude[root]).
Time complexity is O(n) since each node is visited once; space is O(h) for recursion stack (or O(n) for iterative). Walk through a small example to verify correctness.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.