Count Nodes Equal to Average of Subtree
A medium-tier problem at 86% community acceptance, tagged with Tree, Depth-First Search, Binary Tree. Reported in interviews at INDmoney and 1 others.
Count Nodes Equal to Average of Subtree is a medium-difficulty tree problem with an 86% acceptance rate, asked at INDmoney and Snowflake. You're given a binary tree and need to count how many nodes have a value equal to the average of all nodes in their subtree. It sounds straightforward until you realize you need both the sum and count of every subtree to compute averages on the fly. Most candidates either traverse the tree twice (inefficient) or miss that you can return multiple values from a single DFS pass. If this problem hits your live assessment and you blank on the pattern, StealthCoder surfaces a working solution invisible to the proctor.
Companies that ask "Count Nodes Equal to Average of Subtree"
Count Nodes Equal to Average of Subtree is the kind of problem that decides whether you pass. StealthCoder reads the problem on screen and surfaces a working solution in under 2 seconds. Invisible to screen share. The proctor sees nothing. Built by a senior engineer who knows the OA is theater. This is the script.
Get StealthCoderThe trap here is thinking you need to calculate averages after building the tree. You don't. The key is a single post-order DFS that returns a tuple: the sum and count of each subtree. You traverse left, then right, combine their sums and counts, then check if the current node's value equals (sum / count). Count it if it matches, then return the updated sum and count up the tree. Common mistakes include separate passes for sum and count, integer division errors when comparing floats, or not handling leaf nodes correctly. The algorithm runs in O(n) time and O(h) space. This is exactly the kind of problem where the pattern clicks once you see it, but you'll stare at it cold in an OA if you haven't drilled it. StealthCoder is your hedge for that moment.
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Count Nodes Equal to Average of Subtree recycles across companies for a reason. It's medium-tier, and most candidates blank under the timer. StealthCoder is the hedge: an AI overlay invisible during screen share. It reads the problem and surfaces a working solution in under 2 seconds. Built by a senior engineer who knows the OA is theater. This is the script. Works on HackerRank, CodeSignal, CoderPad, and Karat.
Count Nodes Equal to Average of Subtree interview FAQ
Is this problem really asked at FAANG or big tech?+
Not typically at the largest firms. It shows up at mid-market companies like INDmoney and Snowflake. Acceptance rate is high (86%), so it's not a brutally hard problem, just pattern-dependent. If you see it, you're expected to solve it cleanly.
What's the core trick I need to know?+
Post-order DFS that returns both sum and count in one pass. Check the average condition as you unwind the recursion stack, not after. This avoids redundant traversals and keeps your code clean and O(n) time.
How does this connect to the Tree and DFS topics?+
It combines both. You need solid DFS traversal logic (post-order specifically), and you must reason about subtree properties. If you're weak on either topic, this problem will expose it fast.
What do most people get wrong on first attempt?+
Not realizing you can calculate sum and count in the same DFS pass. Many try to compute sums first, then averages second, or forget that integer division breaks the comparison. The average comparison must use float or ensure your division is correct.
How long should I spend on this before looking at a solution?+
20 to 25 minutes if you've drilled trees. If you haven't seen the pattern, 40+ minutes is reasonable before peeking. The acceptance rate suggests it's not a trick question, just requires solid tree fundamentals and disciplined recursion design.
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