Level Order Traversal BFS Binary Tree | Java Fresher Interview
Fast recall
Quick revision
Core answer
Whether you implement BFS with a queue and know the level-size trick to group by level
Trying to do level-order with recursion (that's DFS), or not capturing level boundaries
Use the 'snapshot queue size per level' technique and give O(w) space (widest level)
Interview-ready guide
Interview answer
Level-order traversal visits the tree one level at a time, from the top down, and left to right within each level.
It's really just breadth-first search applied to a tree, and the key tool is a queue, not recursion, because recursion naturally produces a depth-first order instead. The basic algorithm is: enqueue the root, then loop while the queue isn't empty, dequeuing a node, processing it, and enqueuing its non-null children. That alone gives you the nodes in level order.
When the output must be grouped by level as a list of lists, one extra queue-size step marks the boundary. At the start of each iteration of the outer loop, I take a snapshot of the current queue size, call it k. Because of how BFS works, those k nodes currently in the queue are exactly the nodes on the current level.
So I process precisely k nodes, collecting their values into one level's list and enqueuing their children, and when I've done all k, I've finished that level and I move on to the next. The time complexity is O of n, since every node is processed once.
The space complexity is O of w, where w is the maximum width of the tree, the largest number of nodes on any single level, because that's the most the queue ever holds at once, and for a full tree that bottom level can be roughly half the nodes.
The main mistakes to avoid are reaching for recursion, which gives DFS, and forgetting the level-size snapshot, without which you can't tell where one level ends and the next begins.
For example, for root 1, children 2, 3, and grandchildren 4, 5, the queue-size snapshots are 1, then 2, then 2. The grouped result is therefore [[1], [2, 3], [4, 5]].
Level Order Traversal BFS Binary Tree | Java Fresher Interview
Direct Answer
Level-order traversal visits nodes level by level, top to bottom, left to right - it's BFS on a tree. Use a queue: enqueue the root, then repeatedly dequeue a node, process it, and enqueue its non-null children. To group output by level, record the queue size at the start of each level and process exactly that many nodes. O(n) time, O(w) space (w = max width).
Deep Dive & Architecture
Level-order traversal is breadth-first search on a tree. Put the root in a queue, repeatedly remove the front node, and add its non-null children. FIFO order ensures all nodes at one depth are processed before deeper nodes.
Worked example
For root 1, children 2 and 3, and children 4 and 5 under node 2, the queue begins [1]. Remove 1 and add 2, 3, giving [2, 3]. Remove both of those before their children, then the queue holds [4, 5]. The visit order is 1, 2, 3, 4, 5.
To return a list for each level, take a snapshot of queue.size() before processing the level. Process exactly that many nodes into one list; children added during the loop belong to the next level. The snapshots in this example are 1, 2, and 2, producing [[1], [2, 3], [4, 5]].
If the root is null, return an empty result rather than enqueuing null. Every node is enqueued and dequeued once, so time is O(n). Queue space is O(w), where w is the maximum number of nodes on one level.
Code Example
List<List<Integer>> levelOrder(Node root) {
List<List<Integer>> result = new ArrayList<>();
if (root == null) return result;
Queue<Node> q = new ArrayDeque<>();
q.offer(root);
while (!q.isEmpty()) {
int k = q.size(); // snapshot: this level's count
List<Integer> level = new ArrayList<>();
for (int i = 0; i < k; i++) {
Node n = q.poll();
level.add(n.val);
if (n.left != null) q.offer(n.left);
if (n.right != null) q.offer(n.right);
}
result.add(level);
}
return result;
}
// 1 / 2 3 / 4 5 -> [[1],[2,3],[4,5]]
Common Mistakes & Pitfalls
- Using recursion: that's DFS; level-order needs a queue.
- No level-size snapshot: without capturing
q.size()up front, you can't tell where one level ends. - Enqueuing nulls: only enqueue non-null children, or guard before processing.
- Wrong space estimate: space is O(w) (widest level), which can be ~n/2 at the bottom.
Check your understanding
Answer these without looking back at the article.
- What is a binary tree and what are its traversal methods?
- What is a Binary Search Tree (BST) and what are its properties?
- How do you find the height of a binary tree recursively?
Looking for a structured roadmap?
Every module on this page is also part of the full Java backend interview track.