I went with the LCA approach: find the lowest common ancestor, build the path from LCA down to each node, then stitch them together with LCA appearing once in the middle.
Clarify that the path between two nodes goes up from each node to their lowest common ancestor (LCA), then down to the other node. Find the LCA using a recursive or parent-pointer approach, then construct the path by collecting ancestors from each node up to the LCA and concatenating them appropriately.
Pro tip: Mention that if parent pointers are not available, you can find the LCA in a single traversal by returning the node when both target values are found in different subtrees. This shows you understand the trade-offs between different tree representations.
Confirm that the tree is binary, nodes have unique values, and both target values exist. Ask whether parent pointers are available and whether the path should include both endpoints.
Explain that the path between two nodes consists of the upward path from the first node to the LCA and the downward path from the LCA to the second node.
Describe an algorithm to find the LCA, such as recursive post-order traversal that returns the node when both targets are found in different subtrees, or using parent pointers to trace ancestors.
Collect the path from the first node up to the LCA (excluding the LCA), then from the LCA down to the second node (including the LCA). Concatenate these sequences to form the final path.
State the time and space complexity (O(n) time, O(h) space for recursion) and discuss edge cases like one node being an ancestor of the other.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Acknowledge that duplicate node values break assumptions in many tree algorithms, such as BST search or path sum uniqueness. Then systematically discuss how to adapt algorithms by using additional data structures, modifying traversal logic, or redefining the problem to handle duplicates.
Pro tip: Mention that duplicates can cause exponential blowup in path-counting problems, so it's crucial to discuss pruning or memoization to maintain efficiency.
Determine which tree algorithms rely on unique values, such as BST operations, path sum, or LCA. Explain how duplicates invalidate their core assumptions.
For search problems, consider using a multi-set or list to store nodes with the same value, or modify comparisons to handle equality (e.g., go left or right consistently).
For path sum or counting paths, duplicates can create multiple valid paths. Use backtracking with a frequency map or prefix sums to count all possibilities without double-counting.
Discuss time/space trade-offs: e.g., using a hash map to track values vs. sorting nodes. Mention pruning techniques to avoid exploring redundant branches.
Walk through a small tree with duplicate values to demonstrate your approach and ensure correctness, highlighting edge cases like all nodes having the same value.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
First, contrast the generic binary tree LCA approach (O(n) time, O(h) space) with the BST property that allows O(h) time by leveraging node value comparisons. Then, describe the iterative BST LCA algorithm: starting from the root, if both nodes are smaller, go left; if both larger, go right; otherwise, the current node is the LCA. Emphasize the efficiency gain and when it matters.
Pro tip: Mention that the BST property enables an iterative solution with O(1) space (if not counting recursion stack) and that this is a common optimization in production systems where tree height is logarithmic. Also, note that if the tree is not balanced, worst-case time is still O(n), so balancing matters.
Recall that for a generic binary tree, LCA requires traversing all nodes in the worst case, giving O(n) time and O(h) space for recursion.
In a BST, for any node, all values in the left subtree are smaller and all in the right subtree are larger. This allows us to decide which subtree to explore based on node values.
Starting at the root, if both target nodes are less than the current node, move left; if both are greater, move right; otherwise, the current node is the LCA (since the nodes split or one equals the current).
The algorithm visits at most one node per level, so time is O(h) where h is the height of the tree. Space is O(1) if implemented iteratively, or O(h) if recursive.
Mention that for a balanced BST, h = O(log n), so time is O(log n). For an unbalanced BST, h can be O(n), so worst-case time is O(n). Also, handle cases where one node is an ancestor of the other.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.