Binary Tree Level Order Traversal: A Step-by-Step Interview Walkthrough

May 20

10 min read

Binary Tree Level Order Traversal is a problem interviewers like because it forces a specific mental switch. Most tree problems push you toward recursion. This one wants the opposite. The output is grouped by level, which means depth-first traversal won't give you the answer in the right shape without awkward bookkeeping. The interviewer is watching to see whether you recognize that the output format dictates the traversal strategy, and whether you can implement BFS cleanly on a tree. Candidates who default to recursion and then try to massage the output into levels tend to write more code than necessary. Candidates who see "grouped by level" and immediately reach for a queue finish quickly with clean code.

Level-order traversal underpins serialization formats for trees (storing a tree as a sequence that can be reconstructed), shortest-path algorithms on unweighted graphs (BFS from a source visits nodes in order of distance), web crawlers that explore links breadth-first to stay closer to the seed page, UI rendering systems that need to draw parent components before children, and any system that processes hierarchical data tier by tier — like organization charts or dependency graphs.

Problem Statement

Given the root of a binary tree, return the level order traversal of its nodes' values — traversing the tree level by level, from left to right.

Return the result as a list of lists, where each inner list contains the values of the nodes at that level.

Example:

Returns [[3], [9, 20], [15, 7]].

1. Clarify Requirements Before Jumping Into Code

Before writing a single line, we should restate the problem in our own words and ask the kinds of questions that would change our approach.

Input: the root of a binary tree (may be null).

Output: a list of lists of integers, where each inner list contains the values at one level, in left-to-right order.

Details to clarify:

What do we return for an empty tree? An empty list, not a list containing an empty list.
Within a level, is the order strictly left-to-right? Yes.
Are we returning node values or node references? Values.
Can the tree have duplicate values? Yes, but it doesn't affect the algorithm — we never compare values, only positions in the structure.
Can the tree be unbalanced or skewed? Yes.

The thing to flag is that the output shape is what makes this problem different from a generic traversal. We're not just listing nodes — we're grouping them. That grouping is the entire reason BFS is the natural answer, as we'll see in the brainstorm.

2. Identify the Category of the Question

A few signals jump out:

The input is a tree.
The output groups nodes by level.
Within each group, order matters (left to right).

That output structure is the giveaway. Tree problems where the answer is grouped by level — or where we care about "the kth level" or "the farthest level" — almost always want breadth-first search. BFS naturally processes nodes in level order, which is exactly the order we want to produce them. Same family as Binary Tree Right Side View, Average of Levels, and Bottom-Up Level Order Traversal.

3. Brute Force Solution

Let's think about what a naive approach would look like without using BFS. One option:

Compute the maximum depth of the tree.
For each depth d from 0 to maxDepth - 1, traverse the entire tree, collecting nodes whose depth equals d, in left-to-right order.

That works, but it's wasteful. We traverse the entire tree once per level. For a tree of depth h with n nodes, that's O(n × h) time — and we already know we can do this in O(n).

The naive approach reveals the problem with depth-based bucketing: we keep revisiting the same nodes. What we actually want is a single pass that produces the levels in order without ever backtracking. That's exactly what BFS gives us — it processes nodes in level order by construction.

4. Brainstorm More Solutions

Step 1: How would we walk through the example by hand?

Let's trace the example tree:

I'd start at the root and write down 3. Then I'd go to its children: 9, 20. Then I'd go to the grandchildren of the root: 15, 7. The output is exactly [[3], [9, 20], [15, 7]].

Notice the order I visited nodes: root first, then everything at depth 1, then everything at depth 2. That's exactly breadth-first traversal. And the natural way to implement breadth-first traversal is with a queue: enqueue the root, then repeatedly dequeue a node and enqueue its children. When you dequeue, you're visiting nodes in the order you'd encounter them on a level-by-level sweep.

So a queue gets us the order. But the output also needs grouping — we need to know where one level ends and the next begins.

Step 2: How do we know where each level ends?

The queue gives us the right order — nodes come out top-to-bottom, left-to-right. But the problem also requires our output have a grouping. We need to know where the values for level 0 end and the values for level 1 begin. So how do we detect a level boundary?

What if I just dequeue nodes one at a time and watch the queue for some signal? After dequeueing the root, the queue contains [9, 20]. After dequeueing 9, the queue contains [20] (we removed 9 and added nothing because 9 has no children). After dequeueing 20, the queue contains [15, 7] (we removed 20 and added its two children).

Notice what just happened. As I processed the second level, the queue's contents shifted from "all level-1 nodes" to "all level-2 nodes" — but the shift happened gradually, one node at a time. At no single moment did the queue tell me "level 1 is over." There was no marker, no special value, just a smooth transition where level-1 nodes leak out while level-2 nodes flow in.

So the queue alone doesn't tell us where the boundaries are. We need to track the boundaries ourselves. The question is: what information do we have that would tell us where a level ends?

Let's forget about watching the queue mid-level — what if instead we look for a moment when the queue's contents are clean and easy to reason about. Is there ever such a moment?

Yes. Right before we start processing any level, the queue is in a beautifully simple state: it contains exactly the nodes of that level, and nothing else. Every previous level's nodes have already been dequeued (their values are in the output). And no next-level nodes are in the queue yet, because we haven't dequeued any current-level nodes whose children would be added.

That clean moment is the key. If I look at the queue right before I start processing level k, its size is the number of nodes in level k. I can snapshot that number and use it as a counter. Process exactly that many nodes, and I've processed exactly one level. The children I enqueue during those iterations are level-k+1 nodes, but they're added after my snapshot, so they don't inflate the count for the current level.

while queue is not empty:
    levelSize = queue.size()           # snapshot: how many nodes in current level
    level = []
    for i in 0..levelSize - 1:
        node = queue.dequeue()
        level.add(node.value)
        if node.left:  queue.enqueue(node.left)
        if node.right: queue.enqueue(node.right)
    result.add(level)

The snapshot is the entire trick. Without it, we'd lose track of where one level ends and the next begins. With it, the grouping is automatic.

Step 3: Could we do this with DFS instead?

For completeness, BFS isn't the only option. We can solve this with depth-first recursion by passing the current depth as a parameter:

public List<List<Integer>> levelOrder(TreeNode root) {
    List<List<Integer>> result = new ArrayList<>();
    dfs(root, 0, result);
    return result;
}

private void dfs(TreeNode node, int depth, List<List<Integer>> result) {
    if (node == null) return;
    // If this is the first time we're seeing this depth, create its list
    if (depth == result.size()) result.add(new ArrayList<>());
    result.get(depth).add(node.val);
    dfs(node.left, depth + 1, result);
    dfs(node.right, depth + 1, result);
}

This works because we always recurse into the left child before the right child, so within any given depth, values are appended in left-to-right order. The if (depth == result.size()) check lazily creates a new level's list the first time we encounter that depth.

So both DFS and BFS solve the problem in O(n) time. Why prefer BFS?

BFS matches the output structure. The algorithm produces levels one at a time, in the same order the output wants them. DFS produces levels in interleaved order and relies on the result list's structure to do the bucketing.
BFS is easier to reason about. The queue snapshot makes the level boundaries explicit. With DFS, the level boundaries are implicit in the recursion depth, which is less obvious to a reader (and to you, six months later).
BFS extends naturally to follow-ups. Variants like "right side view," "average per level," or "max in each level" plug directly into the BFS loop. With DFS, each variant requires reworking the per-depth bookkeeping.

I'd write the BFS version in an interview and mention DFS as an alternative if the interviewer asks about it.

Step 4: Complexity

The BFS version visits each of the n nodes exactly once and does O(1) work per node (one enqueue, one dequeue, one list append, and at most two more enqueues for children). Total time: O(n).

Space is dominated by the queue. In the worst case — a complete binary tree — the bottom level contains roughly n/2 nodes, all of which sit in the queue at the same time. So queue space is O(n) in the worst case. The output list is also O(n) in total size, but that's required by the problem, not algorithm overhead.

The DFS version is O(n) time and O(h) space for the recursion stack, where h is the height. So DFS uses less space on a balanced tree (O(log n) vs O(n)) but the same space on a skewed tree (O(n) either way). For most realistic inputs, the space difference doesn't matter — both are linear in the input size.

5. Discuss Trade-Offs Between Solutions

Approach	Time	Space	When I'd use it
Multi-pass DFS by depth	O(n × h)	O(h)	Never — wasteful, but shows you understand why BFS is better
BFS with queue	O(n)	O(n)	My default interview answer — matches the output structure naturally
DFS with depth parameter	O(n)	O(h)	A reasonable alternative; mention if asked, but BFS is cleaner

BFS is the clear default for this problem. The grouping in the output and the level-by-level structure of the algorithm align perfectly.

6. Pseudocode

if root is null:
    return empty list

result = empty list
queue = new queue
enqueue root

while queue is not empty:
    levelSize = queue.size()    # snapshot
    level = empty list

    repeat levelSize times:
        node = dequeue
        append node.value to level
        if node.left:  enqueue node.left
        if node.right: enqueue node.right

    append level to result

return result

7. Edge Cases

Things to verify before claiming we're done:

Empty tree (null root) → return empty list immediately. ✓
Single-node tree → one iteration, one level with one value: [[val]]. ✓
Completely skewed tree (all left or all right children) → each level has exactly one node; output is a list of single-element lists. The queue never holds more than one node at a time.
Perfectly balanced tree → standard case; queue holds up to n/2 nodes at the deepest level.
Tree with duplicate values → values are returned as-is; duplicates are fine.

The early null check is the most important edge case to write explicitly. Without it, the algorithm tries to enqueue a null root and breaks.

8. Full Code

import java.util.*;

public class BinaryTreeLevelOrderTraversal {

    static class TreeNode {
        int val;
        TreeNode left;
        TreeNode right;

        TreeNode(int val) {
            this.val = val;
        }
    }

    // BFS solution — O(n) time, O(n) space for the queue.
    // The queue-size snapshot is what gives us level boundaries:
    // before each iteration of the outer loop, the queue contains
    // exactly the nodes of the current level.
    public static List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> result = new ArrayList<>();
        if (root == null) {
            return result;
        }

        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);

        while (!queue.isEmpty()) {
            int levelSize = queue.size(); // snapshot current level
            List<Integer> level = new ArrayList<>();

            for (int i = 0; i < levelSize; i++) {
                TreeNode node = queue.poll();
                level.add(node.val);

                // Children added here belong to the NEXT level,
                // not this one — the snapshot above protects us.
                if (node.left != null)  queue.offer(node.left);
                if (node.right != null) queue.offer(node.right);
            }

            result.add(level);
        }

        return result;
    }

    // DFS alternative
    // Works by passing the current depth and lazily creating a 
    // level list the first time each depth is encountered.
    public static List<List<Integer>> levelOrderDFS(TreeNode root) {
        List<List<Integer>> result = new ArrayList<>();
        dfs(root, 0, result);
        return result;
    }

    private static void dfs(TreeNode node, int depth, List<List<Integer>> result) {
        if (node == null) return;
        if (depth == result.size()) {
            result.add(new ArrayList<>());
        }
        result.get(depth).add(node.val);
        dfs(node.left, depth + 1, result);
        dfs(node.right, depth + 1, result);
    }
}

9. Test the Code

// Standard tree
TreeNode root = new TreeNode(3);
root.left = new TreeNode(9);
root.right = new TreeNode(20);
root.right.left = new TreeNode(15);
root.right.right = new TreeNode(7);
System.out.println(levelOrder(root));
// [[3], [9, 20], [15, 7]]

// Empty tree
System.out.println(levelOrder(null));
// []

// Single node
System.out.println(levelOrder(new TreeNode(1)));
// [[1]]

// Skewed left
TreeNode skewed = new TreeNode(1);
skewed.left = new TreeNode(2);
skewed.left.left = new TreeNode(3);
System.out.println(levelOrder(skewed));
// [[1], [2], [3]]

// Unbalanced — left side deeper than right
TreeNode unbalanced = new TreeNode(1);
unbalanced.left = new TreeNode(2);
unbalanced.right = new TreeNode(3);
unbalanced.left.left = new TreeNode(4);
System.out.println(levelOrder(unbalanced));
// [[1], [2, 3], [4]]

These hit the meaningful cases: a normal tree, the empty case, a single node, a skewed chain, and an unbalanced tree where one side has more depth than the other.

10. Key Lessons

The shape of the output often tells you which traversal to use. When the answer is grouped by level, BFS is the natural fit — it produces levels in the right order without any reshaping. When the answer is "the path from root to some leaf," DFS is the natural fit. Listen to the output shape.
The queue-size snapshot is a small but powerful trick. Whenever you need to process BFS one level at a time — for level grouping, for tracking depth, or for any per-level computation — snapshot the queue size at the start of each outer iteration. The children you add during that iteration belong to the next level.
DFS can solve any tree problem, but BFS often produces cleaner code when the problem cares about levels. Don't force recursion just because trees are usually recursive.
BFS uses O(n) space in the worst case (the bottom level of a balanced tree). DFS uses O(h) space. For large inputs, the choice matters. For most real trees, both are fine.
The queue snapshot pattern is worth remembering. It comes up in this problem and several variants — right-side view, level averages, level zigzag, finding the largest value at each level, and more.

The thing that makes Binary Tree Level Order Traversal click isn't the algorithm — BFS on a tree is mechanical once you've seen it. It's the recognition that the output structure is what determines the traversal strategy. When you train yourself to read output requirements before reaching for a default approach, every other tree problem starts to feel more like a deliberate design choice and less like a reflex.

Good Luck and Happy Coding!