The BST traversal part was fine, I've done that before.
Use the BST property to navigate from the root, keeping track of the closest value seen so far. At each node, compare the node's value to the target, update the closest if needed, and move left if the target is smaller or right if larger. Continue until reaching a null node, then return the closest value.
Pro tip: Explicitly handle the tie-breaking rule by preferring the smaller value when distances are equal, and mention that this approach runs in O(h) time where h is the tree height, which is efficient for balanced BSTs.
Confirm the BST property, the tie-breaking rule (smaller value wins), and that the tree may be empty or have duplicate values. Define 'closest' as minimal absolute difference.
Set a variable to store the closest value found so far, initially the root's value (or None if tree is empty). Also set a pointer to the current node, starting at the root.
While the current node is not null, compare its value to the target. Update the closest value if the current node is closer, or equally close but smaller. Then move left if target < node value, else move right.
After the loop ends, return the closest value. If the tree was empty, return None or handle as appropriate.
State that time complexity is O(h) where h is the height of the tree (O(log n) for balanced, O(n) worst-case), and space complexity is O(1) for iterative approach.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.