The traversal logic itself clicked pretty fast.
Start by clarifying the problem and constraints, then explain the BST property that allows an efficient solution. Present the iterative approach that traverses from the root, moving left or right based on the values of p and q, and finally prove the O(h) time complexity with a clear argument.
Pro tip: Mention that the iterative solution uses O(1) space, which is better than the recursive approach, and discuss how the algorithm handles edge cases like when one node is an ancestor of the other.
Confirm that the tree is a BST, that p and q are guaranteed to exist, and that nodes have unique values. Ask if the tree can be modified or if recursion is preferred.
State that for any node, all values in the left subtree are smaller and all values in the right subtree are larger. This property allows us to decide the direction of traversal based on the values of p and q.
Describe the iterative approach: start at the root, while the current node is not null, if both p and q are smaller, move left; if both are larger, move right; otherwise, the current node is the LCA.
Argue that in the worst case, we traverse from the root to the deepest leaf, visiting at most h nodes, where h is the height of the tree. Thus, time complexity is O(h), which is O(log n) for a balanced BST and O(n) for a skewed tree.
Highlight that the iterative solution uses O(1) extra space. Mention edge cases: p or q is the root, p is an ancestor of q, or the tree is skewed.
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