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Waymo·Software Engineer·Technical Phone Screen·Intermediate

IntermediateRejected
Jul 2026

Summary

Waymo software engineer interview with a tree serialization problem that I did not handle well. Ran out of time before finishing even the basic case, never mind edge cases. Not my best 45 minutes.

Questions Asked (1)

Q1

Given a syntax tree where each node is either a variable or one of the four basic operators (+, -, *, /), serialize the tree into a string using the minimum number of parentheses needed.

Algorithms & Data StructuresTechnical Trade-offs
Author's notes

This wrecked me.

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AI HintsAI Generated

Suggested Approach

Use a recursive traversal that decides whether to parenthesize each child based on operator precedence and associativity. For each node, serialize the left and right subtrees, adding parentheses only when the child's operator has lower precedence than the parent, or when associativity requires it (e.g., right child of '-' or '/').

Pro tip: Mention that you can avoid parentheses for associative operators like '+' and '*' when the child has equal precedence, but must parenthesize the right child for non-associative operators like '-' and '/'. Also, note that parentheses are never needed for leaf nodes.

1. Define precedence and associativity

Assign precedence levels to operators (e.g., +,-: 1; *,/: 2) and specify associativity (left-associative for all four). This determines when parentheses are necessary.

2. Recursive serialization with context

Write a recursive function that takes a node and the parent operator (or context) and returns the serialized string. For each child, decide whether to wrap it in parentheses based on precedence and associativity.

3. Parenthesization rules

Add parentheses if the child's operator has lower precedence than the parent, or if the child is the right operand and has equal precedence with a non-associative parent (e.g., '-' or '/').

4. Handle base cases

For leaf nodes (variables), return the variable name without parentheses. For operators, combine the serialized left and right children with the operator symbol.

5. Test with edge cases

Verify with expressions like a-b-c (should be (a-b)-c without extra parentheses), a-(b-c) (needs parentheses), and mixed precedence like a+b*c.

Key Points to Mention

  • Operator precedence and associativity rules
  • Minimizing parentheses by only adding when necessary
  • Recursive tree traversal
  • Handling left vs. right child differently for non-associative operators
  • Time and space complexity: O(n) time and O(h) space for recursion stack
  • Edge cases: single node, deep trees, mixed operators

AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.