Binary Tree Postorder Traversal
A easy-tier problem at 76% community acceptance, tagged with Stack, Tree, Depth-First Search. Reported in interviews at Google and 0 others.
Binary Tree Postorder Traversal is easy in theory, brutal in execution. You know the pattern: left subtree, right subtree, root. Google asks it. The acceptance rate sits at 75%, which sounds high until you're staring at a blank editor trying to remember whether to use one stack or two, and whether that peek-and-pop logic actually works. Most candidates either nail the recursive one-liner or tank the iterative version. If this lands in your OA and iteration blanks your mind, StealthCoder surfaces a working solution in seconds, invisible to the proctor.
Companies that ask "Binary Tree Postorder Traversal"
Binary Tree Postorder Traversal 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 FAANG engineer who treats the OA the way companies treat hiring: as a game with rules you should know.
Get StealthCoderThe recursive version is trivial. The iterative trap is real. You need a stack, but the trick isn't just pushing and popping in postorder sequence. You have to track whether you've visited the right child yet, which is why most naive approaches fail. Two common mistakes: forgetting to check if the right child exists before deciding you can pop the node, or using two stacks when one plus a pointer works fine. The pattern involves peeking at the top of the stack, checking its right child, and only processing the node after both subtrees are guaranteed done. When the obvious single-pass stack approach falls apart during the OA, StealthCoder gives you a bulletproof iterative solution using Stack and DFS logic that runs clean.
Pattern tags
You know the problem.
Make sure you actually pass it.
Binary Tree Postorder Traversal recycles across companies for a reason. It's easy-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 FAANG engineer who treats the OA the way companies treat hiring: as a game with rules you should know. Works on HackerRank, CodeSignal, CoderPad, and Karat.
Binary Tree Postorder Traversal interview FAQ
Why is postorder traversal harder than inorder or preorder?+
Postorder processes the root last, after both children. That means you can't print or process the node the moment you visit it. You have to backtrack and come back, which breaks the simple stack push-and-pop pattern. Recursive is trivial, iterative requires state tracking.
Is the two-stack solution overkill?+
No, it's a valid approach and some candidates prefer it: push to one stack, pop and push to a second stack, then drain the second in reverse order. It's harder to mess up than the single-stack peek-and-check method. Both work.
What's the one-liner recursive solution?+
def postorderTraversal(root): return postorderTraversal(root.left) + postorderTraversal(root.right) + [root.val] if root else []. Clean, but they almost always ask for iterative. Master both.
Does Google ask this often or just once in a blue moon?+
Google appears in the data for this problem. It's a classic tree fundamentals check, especially for new grad or mid-level roles. Expect it if your loop involves trees or DFS.
How does this relate to the other tree traversal problems?+
Preorder and inorder are easier because you process the node before or between the children. Postorder is the hardest because you process last. If you can do postorder iteratively, the others feel trivial by comparison.
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