Inorder Successor in BST
A medium-tier problem at 51% community acceptance, tagged with Tree, Depth-First Search, Binary Search Tree. Reported in interviews at Pocket Gems and 2 others.
Inorder Successor in BST lands in about half of assessments where it appears, and Pocket Gems, Arista Networks, and Docusign have all asked it. The problem looks straightforward but trips up candidates who don't nail the BST property. You're given a node and need to find the next node in inorder traversal. Most people either brute-force a full traversal or miss that the answer lives in the tree structure itself. If you blank on the pattern during your assessment, StealthCoder solves it in seconds while staying invisible to the proctor.
Companies that ask "Inorder Successor in BST"
Inorder Successor in BST 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 an engineer who got tired of watching his cohort grind for six months and still get filtered at the OA stage.
Get StealthCoderThe trick is recognizing two cases: if the node has a right subtree, the successor is the leftmost node in that subtree. If not, you climb back up the tree to find the first ancestor where your current node is in its left subtree. The gotcha is that many candidates write a traversal first, then optimize. Under live assessment pressure, optimization logic breaks down. The BST property means you don't traverse the whole tree; you follow pointers up and down. DFS thinking is a trap here because you're not exploring; you're navigating. If you hit this problem live and the pattern doesn't click immediately, StealthCoder delivers a working solution in real time without the proctor seeing a thing.
Pattern tags
You know the problem.
Make sure you actually pass it.
Inorder Successor in BST 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 an engineer who got tired of watching his cohort grind for six months and still get filtered at the OA stage. Works on HackerRank, CodeSignal, CoderPad, and Karat.
Inorder Successor in BST interview FAQ
Is Inorder Successor in BST still being asked at top companies?+
Yes. Pocket Gems, Arista Networks, and Docusign have all reported it. It's a medium-difficulty problem with about 50% acceptance, so it's in the sweet spot for screening rounds. Candidates who understand BST navigation tend to solve it fast, which is why it's still in active rotation.
What's the main trick that separates fast solutions from slow ones?+
Recognizing you don't need a full inorder traversal. The BST property lets you navigate directly: right subtree exists, go left from there. No right subtree, climb up until you're the left child of some ancestor. That ancestor is the successor. Full traversal is O(n); this is O(h).
How does this problem relate to general Tree and DFS topics?+
It's filed under both, but DFS (full tree walk) is the trap. The real solution uses BST-specific properties, not general traversal. You climb and search, not recurse downward. Understanding when to ignore the DFS label and use structure instead is the core skill.
What happens if I code a brute-force solution during the assessment?+
You'll likely pass test cases but burn time and risk follow-ups on optimization. The problem is designed to reward BST thinking. If you start writing full traversal code and realize it's inefficient mid-interview, you've lost the momentum edge. StealthCoder hedges that exact scenario.
Is there a common edge case that catches people?+
Yes: when the node has no right child and is the rightmost node in the tree. The successor is null. Also, the given node might not have a parent pointer in some variants, forcing you to search from the root. Check the problem statement carefully before coding.
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