Level Order Traversal —
BFS on a tree, row by row.
Level order visits nodes top to bottom, left to right — one whole row before moving to the next. It's breadth-first search applied to a tree, and it's the only traversal on this hub that needs a queue instead of recursion. Watch each node enqueue its children on LearnBug and see exactly how the queue produces the row-by-row order.
What is Level Order Traversal?
Visit every node at depth 0, then every node at depth 1, then depth 2, and so on — using a FIFO queue. For the tree 4(2(1,3),6) this visits row by row: [4], then [2, 6], then [1, 3]. Unlike inorder/preorder/postorder, this is breadth-first, not depth-first.
Why visualization helps
The tricky part isn't the traversal itself — it's the bookkeeping used to separate levels into their own lists. LearnBug shows the queue's exact contents at every step, so you can see when a level "closes" (the size of the queue at the start of that iteration) instead of guessing at it.
Why this one needs a queue, not recursion
Inorder / Preorder / Postorder — DFS
All three use the call stack (recursion) and go all the way down one branch before backtracking to a sibling.
Level Order — BFS ✓
Uses a FIFO queue instead of the call stack, so it naturally finishes one full row before starting the next.
When level order matters
Anywhere "depth" or "shortest path in an unweighted structure" matters — minimum tree depth, connecting nodes at the same level, serializing by row.
Level order traversal on 4(2(1,3), 6)
Queue starts with just the root — that's level 0
Before the loop starts, the queue holds only [4]. The queue's current size (1) tells the loop exactly how many nodes belong to this level.
Pop 4, enqueue its children 2 and 6 — level 1 forms
The loop runs exactly once for level 0 (queue size was 1). Popping 4 and pushing its two children leaves the queue holding all of level 1.
Pop 2 then 6 — 2 enqueues 1 and 3, node 6 has no children
Queue size was 2 at the start of this level, so the loop runs twice: pop 2 (push 1, 3), pop 6 (nothing to push). Level 2 now sits fully in the queue.
Pop 1 then 3 — both are leaves, queue empties, traversal ends
Neither node 1 nor node 3 has children to enqueue. The queue becomes empty, the while queue loop exits.
Return: [[4], [2, 6], [1, 3]] ✓ every row grouped in visit order
Level order with a queue
def level_order(root):
if root is None:
return []
result, queue = [], [root]
while queue:
level_size = len(queue) # nodes belonging to THIS level
level = []
for _ in range(level_size):
node = queue.pop(0) # dequeue — FIFO
level.append(node.val)
if node.left: queue.append(node.left)
if node.right: queue.append(node.right)
result.append(level)
return resultdef level_order_flat(root):
if root is None:
return []
result, queue = [], [root]
while queue:
node = queue.pop(0)
result.append(node.val)
if node.left: queue.append(node.left)
if node.right: queue.append(node.right)
return result # [4, 2, 6, 1, 3] — same order, one flat listComplexity Notes
Frequently asked questions
Why a queue instead of a stack?
A queue is FIFO — first in, first out — so nodes are processed in the order they were discovered, which naturally finishes an entire level before moving to the next. A stack (LIFO) would jump straight into the most recently discovered branch, producing a depth-first order instead.
How do I know when one level ends and the next begins?
Snapshot len(queue) before the inner loop starts. That number is exactly how many nodes belong to the current level, because every node currently in the queue was enqueued by the previous level — any children pushed during this loop belong to the next level.
Why is list.pop(0) mentioned as a problem?
Removing from the front of a Python list is O(n) because every remaining element has to shift left. For teaching and small trees it's fine, but for large inputs use collections.deque and popleft(), which is O(1).
Is level order traversal the same as BFS on a graph?
Yes — a tree is a special case of a graph with no cycles, so level order traversal is exactly BFS. The only difference is a graph BFS needs a visited set to avoid revisiting nodes; a tree never needs one because there's only one path from the root to any node.
Watch the queue drain one level at a time
Paste your BFS traversal into LearnBug and see the queue's contents update at every step.