How do you find the maximum path sum in a binary tree, where a path can start and end at any nodes?
PICTURE THIS: HOW TO EXPLAIN IT
Simple meaning
DFS returns the best downward gain through one child (or zero if negative).
WHY — Trees instead of guessing?
Why interviewers care about Trees:
question about Trees.
trade-offs, and what you would actually do on a DSA project - not buzzwords.
Name the idea, why it exists, then one short example.
End with when you use it and one common pitfall.
STEPS — What happens step by step?
Before you speak the answer, walk the interviewer through these steps:
- 1DFS returns the best
downward gain through one child (or zero if negative).
- 2At each node update
a global max with node + left_gain + right_gain.
- 3Time is O(n) and
space is O(h).
- 4Returning the through-path instead
of the one-side gain is the usual bug.
- 5Common mistake
What juniors usually get wrong.
- 6Close
When you pick this over the alternative.
EXAMPLE — See it in action
Here's a short line you can speak, broken into clear beats:
Note: Adapt this scaffold to your own project — keep it under 60–90 seconds.
Key takeaway
DFS returns the best downward gain through one child (or zero if negative). At each node update a global max with node + left_gain + right_gain.