High Trees Question 162 of 224

How do you find the maximum path sum in a binary tree, where a path can start and end at any nodes?

DSA interview set · Speak this in 60–90 seconds · Faridabad & Delhi NCR

PICTURE THIS: HOW TO EXPLAIN IT

IdeaTrees
HowWhat happens inside
Why they askShows real use

Simple meaning

DFS returns the best downward gain through one child (or zero if negative).

1

WHY — Trees instead of guessing?

Why interviewers care about Trees:

This is a process

question about Trees.

Panels listen for order,

trade-offs, and what you would actually do on a DSA project - not buzzwords.

Stay structured

Name the idea, why it exists, then one short example.

Close cleanly

End with when you use it and one common pitfall.

2

STEPS — What happens step by step?

Before you speak the answer, walk the interviewer through these steps:

  1. 1
    DFS returns the best

    downward gain through one child (or zero if negative).

  2. 2
    At each node update

    a global max with node + left_gain + right_gain.

  3. 3
    Time is O(n) and

    space is O(h).

  4. 4
    Returning the through-path instead

    of the one-side gain is the usual bug.

  5. 5
    Common mistake

    What juniors usually get wrong.

  6. 6
    Close

    When you pick this over the alternative.

3

EXAMPLE — See it in action

Here's a short line you can speak, broken into clear beats:

Say this line
“Time is O(n) and space is O(h).”
Break into beats
TimeisOnandspace
Speaking order
2987408337471632900

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.

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