https://leetcode.com/problems/binary-tree-level-order-traversal/
Given the root of a binary tree,
Return the level order traversal of its nodes' values (i.e., from left to right, level by level).
-
"Can you restate the problem in your own words?"
→ "We need to traverse a binary tree level by level, collecting all node values at each level from left to right, and return them as a list of lists." -
"What does 'level order' mean exactly?"
→ "We visit all nodes at depth 0 (root), then all nodes at depth 1, then depth 2, etc. Within each level, we go from left to right." -
"What should the output format be?"
→ "A list of lists, where each inner list contains all the values at that level. For example, [[3], [9,20], [15,7]]." -
"What if the tree is empty?"
→ "We should return an empty list []." -
"What if there's only one node?"
→ "We return [[root_value]] - a list containing one list with the root value." -
"How is this different from other tree traversals?"
→ "Unlike inorder/preorder/postorder which use depth-first search, level order uses breadth-first search."
-
"What data structure is typically used for level-by-level processing?"
→ "A queue! It's perfect for breadth-first search since we process nodes in FIFO order." -
"Have you seen similar problems requiring level-by-level processing?"
→ "Yes, like finding the minimum depth of a binary tree, or problems that need to process nodes level by level." -
"How does this relate to graph algorithms?"
→ "This is essentially BFS (Breadth-First Search) applied to a tree structure." -
"What's the key challenge in this problem?"
→ "We need to know when one level ends and the next begins, so we can group the values correctly." -
"Could we use recursion for this?"
→ "Yes, but it's less natural. We'd need to track the current level and build the result differently."
-
"Walk me through the BFS approach with a queue."
→ "Use a queue to store nodes. Process all nodes at current level, add their children to queue for next level, repeat until queue is empty." -
"How do you know when a level is complete?"
→ "Before processing each level, check the current queue size - that's exactly how many nodes are at this level." -
"Could you outline the algorithm step by step?"
→ "1. Handle empty tree. 2. Add root to queue. 3. While queue not empty: get current level size, process that many nodes, collect their values, add children to queue. 4. Return result." -
"What about the recursive approach?"
→ "Pass the current level as a parameter. For each node, ensure the result list has enough levels, then add the node's value to the appropriate level." -
"How would you implement the level-size tracking?"
→ "Before processing each level, record queue.size(). Process exactly that many nodes, which represents the current level."
from collections import deque
class TreeNode:
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
def levelOrder(root: TreeNode) -> List[List[int]]:
if not root:
return []
result = []
queue = deque([root])
while queue:
level_size = len(queue)
current_level = []
# Process all nodes at current level
for _ in range(level_size):
node = queue.popleft()
current_level.append(node.val)
# Add children for next level
if node.left:
queue.append(node.left)
if node.right:
queue.append(node.right)
result.append(current_level)
return resultdef levelOrder(root: TreeNode) -> List[List[int]]:
if not root:
return []
result = []
def dfs(node, level):
if not node:
return
# Ensure we have enough levels in result
if level >= len(result):
result.append([])
# Add current node to its level
result[level].append(node.val)
# Recursively process children
dfs(node.left, level + 1)
dfs(node.right, level + 1)
dfs(root, 0)
return resultdef levelOrder(root: TreeNode) -> List[List[int]]:
if not root:
return []
result = []
current_level = [root]
while current_level:
next_level = []
level_values = []
for node in current_level:
level_values.append(node.val)
if node.left:
next_level.append(node.left)
if node.right:
next_level.append(node.right)
result.append(level_values)
current_level = next_level
return resultdef levelOrder(root: TreeNode) -> List[List[int]]:
if not root:
return []
result = []
queue = deque([root, None]) # None marks end of level
current_level = []
while queue:
node = queue.popleft()
if node is None:
# End of current level
result.append(current_level)
current_level = []
# Add level marker for next level if queue not empty
if queue:
queue.append(None)
else:
current_level.append(node.val)
if node.left:
queue.append(node.left)
if node.right:
queue.append(node.right)
return result-
"What edge cases should we test?"
→ "Empty tree, single node, only left children (skewed left), only right children (skewed right), perfect binary tree." -
"How would you trace through an example?"
→ "For tree [3,9,20,null,null,15,7]: Level 0=[3], Level 1=[9,20], Level 2=[15,7]. Result=[[3],[9,20],[15,7]]." -
"What's the difference between the various approaches?"
→ "BFS with queue is most intuitive and natural. Recursive DFS works but feels less natural. Two queues is clear but uses more space." -
"How do you verify the output is correct?"
→ "Check that each level has the right nodes in left-to-right order, and that all nodes appear exactly once."
def test_levelOrder():
# Empty tree
assert levelOrder(None) == []
# Single node
root = TreeNode(1)
assert levelOrder(root) == [[1]]
# Perfect binary tree
root = TreeNode(3,
left=TreeNode(9),
right=TreeNode(20, TreeNode(15), TreeNode(7))
)
assert levelOrder(root) == [[3], [9, 20], [15, 7]]
# Left skewed tree
root = TreeNode(1,
left=TreeNode(2,
left=TreeNode(3)
)
)
assert levelOrder(root) == [[1], [2], [3]]
# Right skewed tree
root = TreeNode(1,
right=TreeNode(2,
right=TreeNode(3)
)
)
assert levelOrder(root) == [[1], [2], [3]]
# Complex tree
root = TreeNode(1,
left=TreeNode(2, TreeNode(4), TreeNode(5)),
right=TreeNode(3, None, TreeNode(6))
)
assert levelOrder(root) == [[1], [2, 3], [4, 5, 6]]-
"What's the time and space complexity?"
→ "Time: O(n) since we visit each node exactly once. Space: O(w) where w is the maximum width of the tree (for the queue), which is O(n) in worst case." -
"Which approach do you prefer and why?"
→ "BFS with queue is most intuitive since level-order naturally maps to breadth-first search. The recursive approach works but feels forced." -
"How does space complexity vary between approaches?"
→ "Queue-based: O(w) for queue plus O(n) for result. Recursive: O(h) for call stack plus O(n) for result, where h is tree height." -
"What's the maximum width of a binary tree?"
→ "In a perfect binary tree, the last level has about n/2 nodes, so maximum width is O(n). This happens when the tree is complete/nearly complete." -
"Could we optimize space usage?"
→ "If we only needed to process level by level without storing all results, we could use O(w) space. But since we need to return all levels, we need O(n) for the result anyway." -
"How does this compare to other traversal algorithms?"
→ "DFS traversals (inorder, preorder, postorder) use O(h) space and are recursive. BFS uses O(w) space and is naturally iterative." -
"What if we needed to process levels from bottom to top?"
→ "We could use the same algorithm but reverse the result, or use a deque and add levels to the front." -
"What if we needed only specific levels?"
→ "We could modify to stop early when we reach the desired level, or add level filtering logic." -
"How would this work with very large trees?"
→ "Memory usage could be a concern for wide trees. We might need to process in chunks or use streaming approaches for massive datasets."