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The BST property makes this pretty clean once you think about it.
Use the BST property to traverse from the root: if both p and q are less than the current node, move left; if both are greater, move right; otherwise, the current node is the LCA. This yields an O(h) time, O(1) space solution.
Pro tip: Mention that this approach works because the LCA is the first node where p and q diverge in the BST, and note that the iterative version avoids recursion stack overhead, which is often preferred in production code.
Confirm that p and q are guaranteed to exist and are distinct. Discuss edge cases like one node being the ancestor of the other, or the tree being skewed.
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.
Start at the root. While the current node is not null, if both p and q are less than current, go left; if both are greater, go right; otherwise, return current as the LCA.
Time complexity is O(h) where h is the height of the tree (O(log n) for balanced, O(n) worst-case). Space complexity is O(1) for iterative, O(h) for recursive.
Write clean, bug-free code in a language of your choice, handling the traversal and return condition correctly.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Acknowledge that without the BST ordering property, you must traverse the entire tree to find a target or validate properties. Present a recursive or iterative traversal (e.g., DFS or BFS) that visits every node, and analyze time and space complexity. Emphasize that the approach changes from O(log n) to O(n) time, and discuss trade-offs between recursion and iteration.
Pro tip: Mention that for a general binary tree, you can often use the same traversal but must check both subtrees; also note that if the tree is very deep, recursion may cause stack overflow, so an iterative approach with an explicit stack is safer.
Confirm what operation is required (e.g., search, find height, validate BST) and whether the tree is static or dynamic. State that without ordering, no pruning is possible.
Select DFS (preorder, inorder, postorder) or BFS based on the problem. For search, any traversal works; for level-related tasks, BFS is natural.
Describe the recursive or iterative steps: visit node, process it, then recurse/queue left and right children. For iterative DFS, use a stack; for BFS, use a queue.
State that time is O(n) because every node is visited once. Space is O(h) for recursion (h = height) or O(n) worst-case for skewed tree; iterative DFS uses O(h) stack, BFS uses O(w) queue where w is max width.
Compare recursive vs iterative: recursion is simpler but risks stack overflow; iteration is more robust but code is longer. Mention that for very large trees, iterative is preferred.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.