Google Software Engineering Interview: Binary Tree Maximum Path

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Watch our technical coding mock interview with Angie and Simon Ayzman (Software Engineer at Google) as Angie asks, "Given the root of a binary tree of integers, return the maximum path sum."
Simon spends time drawing out a potential tree path to understand the problem. He asks clarifying questions about negative numbers and the structure of the tree overall.
Learn more about Simon here: www.simonayzman.com/
Chapters -
00:00 - Introduction
00:54 - Question
01:07 - Clarifying questions
06:02 - Answer
15:09 - Test cases
24:05 - Interview analysis
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[@tryexponent]

Don't leave your software engineering career path to chance. Make sure you're interview-ready with Exponent's software developer interview prep course. Start free. bit.ly/4eMtImU

[@simonayzman]

Great to be back!

[@devs2530]

I liked the thought process.

[@strawberriesandcream2863]

He’s really good at interviewing! Loved seeing how the solution came together.

[@atanuroy911]

Ah man. I really wish to reach that level of thought process. Feels like I need a mentor.

[@tryexponent]

Check out Exponent's list of expert coaches and mentors to help you think about questions like this! www.tryexponent.com/coaching

[@atanuroy911]

@@tryexponent for free would be great cuz coming from a third world country its not possible to pay that much

[@ThaoPhuong-ln9vc]

@@tryexponent Does Exponent provide courses and solutions in Python only?

[@abubalo]

Great interview

[@harisjalal7575]

Good job Simon!

[@sergejsvisockis]

I wish I could see the times when Google would give a real world coding tasks instead of some kind of unrealistic algebra...

[@lucaspinto2450]

What if they’re all negative? The max 0 only captures some negatives

[@hemanth121]

class Solution { public int maxPathSum(TreeNode root) { this.sol(root); return this.max; } int max = Integer.MIN_VALUE; public int sol(TreeNode root) { if(root == null) { return -(int)1e7; } int sum1 = sol(root.right); int sum2 = sol(root.left); int max1 = Math.max(root.val,Math.max(root.val + sum1,root.val + sum2)); max = Math.max(max1,Math.max(max,root.val + sum1 + sum2)); return max1; } }