Pretty standard BFS question once you realize you just grab the last element of each level's queue snapshot.
Use BFS level-order traversal, processing each level and recording the last node's value. Alternatively, use DFS with a depth parameter, updating the result array when visiting a node at a new depth for the first time (visiting right child before left).
Pro tip: Clarify the definition of 'rightmost node' (e.g., if a level has only a left child, that node is the rightmost). Also, mention that the solution should handle edge cases like an empty tree and that both BFS and DFS are acceptable, but BFS is more intuitive.
Confirm that the right side view is the set of nodes visible when the tree is viewed from the right side, i.e., the rightmost node at each depth. Clarify edge cases: empty tree, skewed tree, etc.
Decide between BFS (level-order traversal) and DFS (pre-order with depth tracking). BFS is straightforward: process each level and take the last node. DFS can be more space-efficient: traverse right-first, and record the first node seen at each depth.
For BFS: use a queue, iterate level by level, and for each level, record the value of the last node. For DFS: use recursion with depth, maintain a result list, and if depth equals result size, append current node's value; recurse right then left.
Walk through a sample tree (e.g., [1,2,3,null,5,null,4]) to verify the output [1,3,4]. Also test edge cases: empty tree, single node, left-skewed tree.
State time complexity O(n) and space complexity O(n) for BFS (queue) or O(h) for DFS (recursion stack), where h is tree height. Mention that both are optimal.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Same structure, just grab index 0 instead of the last.
Clarify that this is a variation of level-order traversal where we record the first node at each level. Use BFS with a queue, and for each level, capture the first node's value before processing the rest. Alternatively, use DFS with depth tracking, updating the result only when visiting a level for the first time.
Pro tip: Mention that BFS is more intuitive for level-based problems, but DFS can be more space-efficient for skewed trees. Also, handle edge cases like an empty tree and a tree with only right children.
Confirm that the left side view means the leftmost node at each depth, and that we return a list of values from top to bottom.
Decide between BFS (queue) and DFS (recursion/stack). BFS naturally processes level by level; DFS can track depth and update the result when a new depth is reached.
For BFS: enqueue root, then for each level, record the first node's value and enqueue children. For DFS: recurse with depth, and if depth equals result size, append current node's value.
Check for null root and return an empty list. Ensure the algorithm works for trees with only right children (leftmost node is the right child).
State time complexity O(n) and space complexity O(w) for BFS (w = max width) or O(h) for DFS (h = height).
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Still BFS, just sum the level and divide by count.
Use a breadth-first search (BFS) with a queue to traverse the tree level by level. For each level, compute the sum of node values and divide by the number of nodes to get the average. Collect these averages in a list and return it.
Pro tip: Clarify upfront whether the tree can be empty and how to handle division by zero (e.g., return an empty list). Also, mention that using a queue with level-size tracking avoids needing to store all nodes, keeping space complexity O(w) where w is the maximum width.
Ask about input constraints: can the tree be empty? What should be returned for an empty tree? Are node values within a certain range? Confirm the expected output format.
Explain that BFS naturally processes nodes level by level. Use a queue to store nodes of the current level, and track the number of nodes at that level to compute the average.
While the queue is not empty, record the current level size, iterate that many times, summing node values and enqueuing children. After the loop, compute the average and add to the result list.
State time complexity O(n) since each node is visited once, and space complexity O(w) where w is the maximum width of the tree. Walk through a small example to verify correctness.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.