DSA interview question

Lowest Common Ancestor in a Binary Tree

Hard · Data structures · Topics: tree, dfs, recursion

The problem

A company org chart is a binary tree, and HR needs the lowest-ranking manager that both of two employees report up to: their lowest common ancestor (LCA), where a node can be an ancestor of itself. In a plain binary tree there's no ordering to steer the search, so do one post-order DFS. Each call returns p or q if it finds one in its subtree (or the LCA if it finds both). The first node where p and q come back from different sides is the answer.

Example: tree [3, 5, 1, 6, 2, 0, 8, null, null, 7, 4], p = 5, q = 1 → 3; p = 5, q = 4 → 5.

Function signature

TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q);  // p and q exist in the tree

Test cases

Input `(tree, p, q)`	Output	Explanation
`([3,5,1,6,2,0,8,null,null,7,4], 5, 1)`	`3`	On opposite sides of the root
`([3,5,1,6,2,0,8,null,null,7,4], 5, 4)`	`5`	A node is its own ancestor
`([3,5,1,6,2,0,8,null,null,7,4], 6, 4)`	`5`
`([3,5,1,6,2,0,8,null,null,7,4], 7, 8)`	`3`
`([1, 2], 1, 2)`	`1`

(Confirm: values are unique; both p and q are in the tree; a node may be its own ancestor.)

How interviewers phrase it

"Given a binary tree, find the lowest common ancestor of two given nodes" (LeetCode 236).

Clarifying questions to ask

Are both nodes guaranteed to be in the tree? (Yes. Otherwise, count how many were actually found.)
Can a node be its own ancestor? (Yes.)
Is it a BST? (No. For a BST, walk down using values.)

Follow-ups to expect

"Why can the function return early when it finds p or q?"
"How do you know the LCA was found?"
"Alternative with parent pointers?"
"Many queries?"
"Complexity?"

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