Flip Equivalent Binary Trees
A medium-tier problem at 70% community acceptance, tagged with Tree, Depth-First Search, Binary Tree. Reported in interviews at Anduril and 0 others.
Flip Equivalent Binary Trees is a medium-tier tree problem that appears in Anduril's assessment pipeline. You're given two binary trees and need to determine if one is a flipped version of the other, where flipped means left and right children are swapped at some nodes. The 70% acceptance rate masks the deception: candidates either spot the recursive pattern immediately or waste 15 minutes on the wrong approach. This problem tests whether you understand how flipping propagates through a tree structure and when two trees can be equivalent despite different child orderings.
Companies that ask "Flip Equivalent Binary Trees"
Flip Equivalent Binary Trees 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. Made by a working Amazon engineer who got tired of watching qualified friends bomb OAs they'd solve cold in an IDE.
Get StealthCoderThe trick is recognizing that two trees are flip-equivalent if they match structurally after flipping some subtrees. The recursive pattern: two nodes match if their values are equal and either (1) left child of tree1 matches left child of tree2 and right matches right, or (2) left child of tree1 matches right child of tree2 and right matches left. Most candidates start by comparing nodes directly without accounting for the flip dimension, then scramble when test cases fail. The naive approach assumes fixed parent-child relationships. The actual solution explores both matching possibilities at each node using DFS. If you hit this live during your assessment and can't immediately articulate the two-case recursion, StealthCoder surfaces the pattern in seconds, invisible to the proctor, so you don't crater.
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Flip Equivalent Binary Trees 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. Made by a working Amazon engineer who got tired of watching qualified friends bomb OAs they'd solve cold in an IDE. Works on HackerRank, CodeSignal, CoderPad, and Karat.
Flip Equivalent Binary Trees interview FAQ
Is this about whether trees are identical, or something else?+
No. Identical trees have the same structure and values in the same positions. Flip-equivalent trees can have left and right children swapped at any nodes. You're checking if one tree can become the other through a sequence of subtree flips, which is a harder structural comparison.
Why do candidates struggle with this one even at 70% acceptance?+
Because the problem description doesn't explicitly say you must try both orientations at each node. Candidates compare node-by-node as if the tree is fixed, then hit a wall. The recursive insight that both matching patterns are valid at every step takes a mental shift.
What's the time complexity, and does it matter for Anduril?+
O(min(n, m)) where n and m are tree sizes, because you traverse both trees once, branching into two recursive calls. Standard assessment constraints won't penalize this. Space is O(height) for the call stack.
How does this relate to the Tree and DFS topics listed?+
You're using DFS to traverse both trees simultaneously, and the recursion naturally follows the tree structure. It's a pure recursive tree traversal problem that combines node values with structural comparison. No graphs, no iterative tricks required.
Should I implement this iteratively or recursively?+
Recursively. The problem is designed for recursive thinking because the flip equivalence is defined recursively. An iterative solution with queues adds complexity and cognitive load without benefit in a live assessment.
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