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My first instinct was to treat it like a generic tree and do some kind of recursive scan, which would've worked but completely missed the point.
Start by clarifying the problem and assumptions, then explain the BST property that allows an efficient solution. Walk through the iterative approach that traverses from the root, moving left or right based on the values of p and q, and finally return the node where they split.
Pro tip: Mention that the BST property enables O(h) time and O(1) space, which is optimal. Also, briefly discuss how the solution changes if the tree is not a BST, showing depth of understanding.
Confirm that the tree is a BST, nodes p and q are guaranteed to exist, and we need the lowest common ancestor (LCA). Ask if the tree can be modified or if we can use extra space.
State that for any node, all values in the left subtree are smaller and all values in the right subtree are larger. This allows us to decide the direction of traversal based on the values of p and q.
Start from the root. While the current node is not null, if both p and q are smaller than the current node, move to the left child; if both are larger, move to the right child; otherwise, the current node is the LCA.
Time complexity is O(h) where h is the height of the tree, and space complexity is O(1) for the iterative approach. Mention that recursion would use O(h) space.
Discuss cases where p or q is the root, or one is an ancestor of the other. Also, briefly mention how to solve it if the tree were not a BST (e.g., using recursion or parent pointers).
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