Diameter of Binary Tree || Leetcode 543 || 1 Variant Question Big Tech Actually Asks

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Discover the actual variant Meta asks on Leetcode problem 543: Diameter of Binary Tree.

Timestamps:
00:00 Leetcode Explanation
12:41 Leetcode Coding
14:10 Variant Explanation: N-ary Tree
21:08 Variant Coding: N-ary Tree

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FAANG, mid-sized companies, and startups are asking more LeetCode-style puzzle questions every day, making it harder to stand out as the competition grows. With an increasing number of new graduates entering the software market and tech companies laying off developers while overworking those who remain, it’s a tough landscape. Take Meta, for example: they expect 2-3 months of intense study time, only to likely ghost you afterward. But this doesn’t mean we should be unprepared.

While LeetCode is a valuable learning resource, many developers focus too much on rote memorization. Others find themselves stuck in a vicious cycle, where they don’t study as efficiently as they could because they’re juggling multiple responsibilities. They have full-time jobs, personal commitments, or other obligations that limit the time they can dedicate to solving problems. It’s a grind. Unfortunately, most companies introduce their own twists or "variants" of common problems (e.g., 6-sum instead of 2-sum), which throw candidates off. Rephrasings of problems and follow-up questions are also common, so recognizing these variations and curveballs is crucial.

For those who don’t have the time to revisit LeetCode problems multiple times to solidify concepts, this channel covers the most frequently asked variants, rephrasings, and follow-ups. If you've seen these before, you’ll have a significant edge over your competitors. Remember, time pressure — especially at Meta — is intense, so speed is essential. Even with thorough preparation, interviewers may be unpredictable, but knowing the variants beforehand can drastically increase your chances of success.

Take LeetCode 543, Diameter of Binary Tree, which is one of Meta’s most frequently asked questions (top 10 as of writing). Meta mostly sticks to the original problem. Variants - however - do exist, and with thousands of interviewers, it's hard to predict them all. We cover a key variant: what if you had to return the diameter of an N-ary tree?

We also walk through variants in mock interviews. Using the exact platform (CoderPad) that Meta uses, you’ll get familiar with the UI, settings, and overall experience. This is a 1-on-1 simulation of how Meta conducts and facilitates their interviews, so the goal is to avoid wasting time on the tool itself. Once again, time limits at Meta are a huge factor, so speed is critical.

That said, even with perfect preparation, interviewers might not always be in a good mood or may judge unfairly. The interview process has its power dynamics, but with insider knowledge, you’ll have done as much as you can on your end to secure a Strong Hire decision.
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An important lesson to note here, that we are never returning the diameter as returning the diameter won't help us in any way in the recursive stack, and so we always return just the max height in any case.

adityajindal
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I think the best idea is to use a Priority Queue with size 2 instead of busting the brains with two variables - TC/SC wont change.

arunraj
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A question in the n-ary variant. Would it look a lot cleaner if in the helper function, we just maintain a minHeap of length 2 and simply add all the returned diameters in there and then we just have the 2 max diameters to work with? I know it will add another O(klog2) time per function call (k being the average num of children) but would it be worth it to mention it at least? Thoughts?

TravelGrid
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Hey Minmer! Do you guys offer mock interview services?

TheMrmancave
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Question: For N-arry tree, will the tree structure be given by the interviewer or we need to draw it by self from the input

Neha-bdk
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For both problems why the space complexity is O(N)? if the diameter is a pointer and we use only 2-3 other int vars to store and calculate the length? shouldn't be O(1)?

paulcruz
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At 349 you say it os n^2, how? In order to do dfs you must traverse to leaf and then return up through callstack. You are not repeatedly going through every node n times. You only visit nodes 1 time so o(n)

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