Lowest Common Ancestor of a Binary Search Tree
A medium-tier problem at 68% community acceptance, tagged with Tree, Depth-First Search, Binary Search Tree. Reported in interviews at X and 7 others.
You're in the OA, trees are live, and you're staring at a binary search tree with two nodes. Find their lowest common ancestor. Sounds simple until you realize there are multiple valid approaches and you've got seconds to pick the right one. This problem shows up across X, LinkedIn, Google, Meta, Microsoft, and other major tech companies. The acceptance rate sits at 68%, which means most people who see it solve it, but that's after they've either drilled it or thought through the BST property carefully. If you blank on the pattern during your live assessment, StealthCoder surfaces a clean solution in seconds, invisible to the proctor.
Companies that ask "Lowest Common Ancestor of a Binary Search Tree"
Lowest Common Ancestor of a Binary Search Tree 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 trap is trying to solve this like a regular binary tree, running DFS on both subtrees and merging results. Wrong move. The trick is the binary search tree property itself: values in the left subtree are smaller, values in the right subtree are larger. Once you see that, the solution becomes linear time with no recursion overhead. Start at the root. If both target nodes are smaller than the current node, go left. If both are larger, go right. If they're on opposite sides or one is the current node, you've found the LCA. Three lines of code, no tree traversal needed. The problem tests whether you actually understand BST properties or just pattern-match tree problems. During a live OA, if the obvious recursion doesn't feel right, StealthCoder gives you the iterative BST solution fast so you don't waste time second-guessing.
Pattern tags
You know the problem.
Make sure you actually pass it.
Lowest Common Ancestor of a Binary Search Tree 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.
Lowest Common Ancestor of a Binary Search Tree interview FAQ
Is this problem actually easier than it sounds?+
Yes, if you use the BST property. Most candidates overthink it and code a general tree LCA solution, which works but is slower. The BST property lets you cut the tree in half at each step, making it trivial. Acceptance at 68% confirms most people figure it out, but many take the long route.
Do I need recursion for this?+
No. Recursion works but it's overkill. The iterative approach is cleaner: traverse from root, compare values, and move left or right based on BST ordering. Same time complexity, less call-stack overhead. Either passes, but iterative is sharper.
What if one of the target nodes isn't in the tree?+
The problem assumes both nodes exist in the tree. If the OA variant doesn't explicitly state that, clarify with the proctor. Standard LeetCode version guarantees both are present. You can add a validation step if unsure, but read the spec first.
How does this relate to Depth-First Search?+
DFS is the general backup if you don't catch the BST trick. You traverse the tree fully, searching both subtrees, then merge results. It works for any binary tree but wastes time here. The BST property eliminates the need for full DFS, which is why the problem is MEDIUM, not HARD.
Is this still asked at major tech companies?+
Yes. Google, Meta, Microsoft, LinkedIn, and Oracle all report asking it. It's popular because it's medium-difficulty and tests whether you think algorithmically about properties, not just code. Shows up less frequently than classic array problems but often enough to matter for prep.
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