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Pattern1 Practice1 Interview

Tree DP

Aggregate dynamic-programming states from child subtrees into a result for each parent node. Learn its decision rule, maintained state, Python cost model, and transfer from a focused drill to a full Interview Problem.

Recognize it when

Consider Tree DP when the prompt's constraints and required operations match this shape: Aggregate dynamic-programming states from child subtrees into a result for each parent node.

Pybit demonstrates the Tree DP decision pattern in a professional coding interview workspace.
On this page · Return a Summary from Each Subtree

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Tree DP Code Labs

Return a Summary from Each Subtree

Tree DP defines what one node returns after solving its entire subtree. The summary may contain multiple modes because the parent needs different answers under different choices.

Combine Children in Postorder

Solve every child before the parent, then combine independent child summaries. The recursion frame owns the parent-child relation and prevents using an unfinished subtree.

Trace Tree DP Summaries

Reference
def subtree_size_trace(children,root):
    trace=[]
    def solve(node):
        size=1+sum(solve(child) for child in children[node]);trace.append([node,size]);return size
    if root!=-1:solve(root)
    return trace
Practice

Implement subtree_size_trace(children,root). Return [node,size] when each node completes in postorder.

Public tests

  • Combine completed child summaries

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Keep Local and Global Answers Separate

A returned path or state may need to extend through the parent, while the best answer anywhere in the subtree may combine multiple children. Track these meanings separately rather than returning an invalid shape upward.

Maximum Independent Tree Sum

Reference
def max_independent_sum(values,children,root):
    if root==-1:return 0
    def solve(node):
        take=values[node];skip=0
        for child in children[node]:
            child_take,child_skip=solve(child);take+=child_skip;skip+=max(child_take,child_skip)
        return take,skip
    return max(solve(root))
Practice

Implement max_independent_sum(values,children,root). Choose maximum node-value sum with no parent-child pair both chosen.

Public tests

  • Return take and skip per subtree

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Choose Tree DP or Rerooting

Use ordinary tree DP when one root and child-to-parent summaries suffice. Use rerooting when the answer is required for every possible root, combining downward and parent-side contributions in two passes.

Explain It in an Interview

Say: “Each call returns these modes for its subtree. After all children complete, I combine their compatible states and return only information the parent can legally extend.” Count O(nodes × state combinations) time and O(height) stack.

Python Version Note

Python 3.11+

The examples use the standard Python syntax and containers supported by the browser Judge; the Tree DP invariant is independent of a minor Python release.

Verify in Python docs(opens in a new tab)

Complexity & Invariants

Interview bounds depend on the stated representation, input model, and real Python operations.

OperationAverageWorstInterview note
Tree DP decision loopO(n)O(n)Build child adjacency and return the optimal take and skip totals from each subtree.

Space

O(n) for the focused Choose Non-Adjacent Tree Values implementation.

Assumptions

  • When a node is taken, every child must be skipped; when it is skipped, each child independently chooses its better valid state. These cases partition all legal selections, so the better root state is globally optimal.
  • The bound includes real Python operations; no repeated slicing, linear list membership, or hidden sorting is treated as free.

Invariants Worth Saying Aloud

  1. State the precise Tree DP invariant before coding and preserve it after every update, traversal step, or recursive return.
  2. A returned state summarizes the entire subtree under one explicit root condition.
  3. Sibling subtrees combine independently once the parent decision is fixed.
  4. Define take and skip states for each subtree. Preserve this property after every transition.
  5. Combine independent child choices under the parent decision. Preserve this property after every transition.

When To Use Or Avoid Tree DP

Use It When

  • Use Tree DP when the prompt exposes the ordering, monotonicity, recurrence, or constraint that makes this decision safe.
  • Use it when a returned state summarizes the entire subtree under one explicit root condition.

Choose Another Tool When

  • Avoid Tree DP when its decision rule cannot eliminate work or preserve the required state.
  • Avoid forcing the label onto a prompt whose simpler direct scan already meets the constraints.

Common Pitfalls

These are the mistakes most likely to survive a happy-path example and fail a boundary case.

Moving state without a proof

Applying the pattern without proving its ordering, monotonicity, or window invariant can silently miss valid candidates.

Prevent it: Write the decision rule beside the loop and verify it against Choose Non-Adjacent Tree Values before optimizing.

Avoid

def maximum_tree_independent_sum(parents, values):
    pass

Use instead

def maximum_tree_independent_sum(parents, values):
    if not values:
        return 0
    children = [[] for _ in values]
    for node in range(1, len(values)):
        children[parents[node]].append(node)
    def solve(node):
        take = values[node]
        skip = 0
        for child in children[node]:
            child_take, child_skip = solve(child)
            take += child_skip
            skip += max(child_take, child_skip)
        return take, skip
    return max(solve(0))

Where you will hit this: Choose Non-Adjacent Tree Values(opens in a new tab)

Breaking the maintained state

Always takes each node together with the best child state, allowing adjacent selections.

Prevent it: Use the public tests and preserve this state: A returned state summarizes the entire subtree under one explicit root condition.

Avoid

def maximum_tree_independent_sum(parents,values):
    return sum(values)

Use instead

def maximum_tree_independent_sum(parents, values):
    if not values:
        return 0
    children = [[] for _ in values]
    for node in range(1, len(values)):
        children[parents[node]].append(node)
    def solve(node):
        take = values[node]
        skip = 0
        for child in children[node]:
            child_take, child_skip = solve(child)
            take += child_skip
            skip += max(child_take, child_skip)
        return take, skip
    return max(solve(0))

Where you will hit this: Choose Non-Adjacent Tree Values(opens in a new tab)

Hiding Python work in the hot path

Python slicing, copying, membership in lists, or repeated sorting inside the loop can invalidate the intended complexity.

Prevent it: Count every slice, copy, sort, membership check, and container update before claiming O(n).

Avoid

def maximum_tree_independent_sum(parents,values):
    return sum(values)

Use instead

def maximum_tree_independent_sum(parents, values):
    if not values:
        return 0
    children = [[] for _ in values]
    for node in range(1, len(values)):
        children[parents[node]].append(node)
    def solve(node):
        take = values[node]
        skip = 0
        for child in children[node]:
            child_take, child_skip = solve(child)
            take += child_skip
            skip += max(child_take, child_skip)
        return take, skip
    return max(solve(0))

Where you will hit this: Maximum Contiguous Subarray Sum(opens in a new tab)

Reviewed References

Official Python documentation supports language behavior. Canonical problem pages provide additional practice context.