No. That combination violates the adjacency rule. Return two answers from each child: its best value when its root is taken, and when its root is skipped.

Represent nodes as (value,left,right). Let solve(node) return (take,skip). Taking forces both children to be skipped; skipping lets each child choose its better case. Picking nothing is allowed.

def tree_pick(root):
    def solve(node):
        if node is None:
            return 0, 0
        value, left, right = node
        lt, ls = solve(left)
        rt, rs = solve(right)
        take = value + ls + rs
        skip = max(lt, ls) + max(rt, rs)
        return take, skip
    return max(solve(root))

assert tree_pick((5, (4, None, None), (4, None, None))) == 8
assert tree_pick((-5, None, None)) == 0
assert tree_pick(None) == 0

After conditioning on whether a node is selected, its child subtrees are independent under these rules. That explains why their contributions can be added. Each node computes two values once: O(n) time and O(h) recursion space, without needing a separate memoization dictionary on a proper tree.

This is a useful correction to “DP means adding a cache.” The important feature here is the conditional subproblem values and recurrence; the traversal already ensures each subproblem is evaluated once.

For a general graph with cross-edges between branches, those supposedly separate subtrees can constrain each other. Applying the same recurrence would ignore those edges. The tree structure makes the small boundary summary sufficient.

Analogy limit: two return values do not automatically mean “2D DP” in the usual table sense. One can describe the logical state as (node, selected_status), but implementation and naming conventions vary; explain its information rather than defending a label.

No. Selections in different child branches can now conflict through their parent. The boundary summary needs enough distance information to check those interactions before combining children.