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Start by clarifying the definition of maximum depth and edge cases, then present a recursive DFS solution that computes the depth as 1 + max(depth(left), depth(right)). Discuss the time and space complexity, and optionally mention an iterative BFS alternative.
Pro tip: Show awareness of potential stack overflow in recursion for very deep trees and suggest an iterative approach as a follow-up, demonstrating production-level thinking.
Confirm that maximum depth is the number of nodes along the longest path from root to leaf, and discuss edge cases like empty tree (depth 0) and single node (depth 1).
Decide between recursive DFS (simpler, elegant) and iterative BFS (avoids recursion limits). Explain your choice based on constraints and clarity.
Write clean code for the chosen approach. For DFS: if root is null return 0; else return 1 + max(maxDepth(left), maxDepth(right)). For BFS: use a queue and count levels.
State that time complexity is O(n) since each node is visited once, and space complexity is O(h) for DFS (h = height) or O(w) for BFS (w = max width).
Walk through a small example (e.g., [3,9,20,null,null,15,7] returns 3) to verify correctness and edge cases.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Use an iterative BFS with a queue to process nodes level by level, capturing each level's values before moving to the next. Clearly explain the algorithm, then implement it with attention to edge cases and complexity.
Pro tip: Mention that you can use a sentinel (e.g., null) or track the queue size to separate levels, and note that this approach is easily adaptable to variations like zigzag traversal.
Restate the problem to ensure understanding: return a list of lists where each inner list contains the values at that depth. Ask about edge cases like empty tree or skewed tree.
Explain that level-order traversal naturally uses BFS. Use a queue to process nodes level by level, and for each level, record the values of all nodes at that depth.
Initialize a queue with the root. While the queue is not empty, determine the current level size, dequeue that many nodes, collect their values, and enqueue their children. Append the level list to the result.
State that time complexity is O(n) since each node is visited once, and space complexity is O(m) where m is the maximum number of nodes at any level (or O(n) in the worst case).
Walk through a simple example (e.g., [3,9,20,null,null,15,7]) to verify the output. Mention edge cases like empty tree (return []) and single node (return [[root.val]]).
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Use a depth-first search (DFS) to traverse all root-to-leaf paths, constructing the string for each path by prepending the current node's character as you go. Keep track of the lexicographically smallest string found so far, comparing at each leaf. Alternatively, use a recursive function that returns the smallest string from a subtree, combining the current node's character with the smallest child string.
Pro tip: Mention that you can optimize by pruning branches: if the current path's prefix is already lexicographically larger than the best found so far, you can stop exploring that branch. This shows awareness of efficiency beyond the naive approach.
Confirm that the string is formed by reading from leaf to root, so the leaf's character is the first character of the string. Discuss edge cases like a single-node tree (root is leaf) and trees with varying depths.
Decide between top-down DFS (building strings as you go down) or bottom-up recursion (returning smallest string from children). Explain why DFS is suitable since we need to explore all paths.
For bottom-up: at each node, recursively get the smallest string from left and right subtrees, then prepend the current node's character to the smaller one. For top-down: pass the current string (reversed) and update the global minimum at leaves.
At a leaf, the string is just the leaf's character. When comparing strings, use lexicographic order. Ensure that if one subtree is null, you only consider the other.
Time complexity is O(N * L) where N is number of nodes and L is max depth (due to string concatenation/comparison). Space is O(H) for recursion stack. Mention pruning or using a trie-like approach for optimization.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.