Level Order Traversal

1. One-Liner

Level order (BFS) visits a binary tree row by row, left to right, using a queue FIFO.

2. The Problem It Solves

You need breadth-first view: each depth together, shortest path on an unweighted tree graph, or building level-linked lists. It is the tree specialization of BFS.

3. The Core Idea

Enqueue the root; repeatedly dequeue the front, record its value, enqueue left then right children. Process all nodes at depth d before any at d+1 by processing the queue in size-sized chunks per level.

4. How It Works (Step-by-Step)

Step	Action
1	Queue starts with root.
2	While queue not empty: let `k = queue.size()`.
3	Dequeue `k` times → one level; enqueue children.
4	Append level values to answer.
5	Continue until no nodes left.

5. Dry Run Example

Tree: root 1, children 2,3, 2’s children 4,5.

Level	Queue snapshot	Output row
0	[1]	[1]
1	[2,3]	[2,3]
2	[4,5]	[4,5]

6. Key Properties

Property	Detail
Order	By increasing depth
Queue	Essential; stack gives DFS
Zigzag	Alternate reversal per level

7. Where It Is Used

Area	Use
UI / org charts	Hierarchy by rank
Networking	Broadcast trees
Interviews	Serialize by level, width of tree

8. Interview Tips

O(n) time is standard; space O(w) where w = max width (up to n/2 near bottom). Handle null in queue only if pattern requires (e.g., complete tree check).

9. Comparison with Other Algorithms

Traversal	Data structure	Order
Level order	Queue	Breadth
Preorder DFS	Stack/recursion	Depth
Vertical order	HashMap + BFS	Column

10. Complexity


--	--
Time	O(n)
Space	O(w) queue, w ≤ n

Implementation Example (PYTHON)

from collections import deque

def level_order(root):
    if not root:
        return []
    q, ans = deque([root]), []
    while q:
        level = []
        for _ in range(len(q)):
            n = q.popleft()
            level.append(n.val)
            if n.left: q.append(n.left)
            if n.right: q.append(n.right)
        ans.append(level)
    return ans