Binary Tree Level Order Traversal: LeetCode 102 Walkthrough

If you are going to learn one BFS tree problem, make it LeetCode 102. Binary tree level order traversal is the foundation for nearly every level-based tree question you will encounter in coding interviews. The pattern you learn here transfers directly to at least half a dozen other tree problems.

The idea is straightforward: given a binary tree, return its nodes grouped by level, from top to bottom, left to right. The output is a list of lists — each inner list contains the values at one level of the tree. What makes this problem valuable is not difficulty but the template it teaches.

Once you internalize the BFS queue pattern from this problem, you can solve Zigzag Level Order Traversal (#103), Binary Tree Right Side View (#199), and Average of Levels in Binary Tree (#637) with minor modifications. That is why interviewers love it — it tests whether you truly understand BFS on trees.

The problem gives you the root of a binary tree and asks you to return the level order traversal as a list of lists. Each sublist contains the values of nodes at that depth, ordered left to right. For example, given the tree [3,9,20,null,null,15,7], the expected output is [[3],[9,20],[15,7]].

Level 0 has just the root node with value 3. Level 1 has nodes 9 and 20, the children of the root. Level 2 has nodes 15 and 7, the children of node 20. Node 9 has no children, so it contributes nothing to level 2. The output captures this exact structure.

This is fundamentally a breadth-first search problem. You need to visit every node at depth d before visiting any node at depth d+1. That requirement maps perfectly to a queue data structure, which processes elements in first-in-first-out order.

The BFS level order approach uses a queue to process nodes one level at a time. The key insight is tracking how many nodes belong to the current level before you start processing them. At the beginning of each level, the queue contains exactly the nodes for that level — its length tells you how many to dequeue.

You dequeue that exact number of nodes, collecting their values into a level list. As you process each node, you enqueue its left and right children if they exist. By the time you have processed all nodes for the current level, the queue contains exactly the nodes for the next level. Then you repeat.

This size-tracking trick is what separates a simple BFS traversal from a level-aware BFS traversal. Without it, you would visit nodes in the right order but lose track of where one level ends and the next begins. With it, you get clean level boundaries every time.

The implementation follows the BFS template closely. You start by handling the edge case of an empty tree — if root is null, return an empty list. Otherwise, initialize a queue with the root node and an empty result list.

The outer loop runs while the queue is not empty. At the start of each iteration, record the current queue size — this is the number of nodes at the current level. Create an empty list for this level. Then run an inner loop exactly that many times.

In the inner loop, dequeue the front node, add its value to the current level list, and enqueue its left and right children if they are not null. After the inner loop completes, append the level list to the result. When the outer loop finishes, return the result.

Let us walk through the classic example: the tree [3,9,20,null,null,15,7]. The root is 3, its left child is 9, its right child is 20. Node 9 has no children. Node 20 has left child 15 and right child 7.

Start: queue = [3]. Level 0: levelSize = 1. Dequeue 3, add to level list. Enqueue children 9 and 20. Level list = [3]. Result = [[3]].

Queue = [9, 20]. Level 1: levelSize = 2. Dequeue 9, add to level list, no children to enqueue. Dequeue 20, add to level list, enqueue 15 and 7. Level list = [9, 20]. Result = [[3], [9, 20]].

Queue = [15, 7]. Level 2: levelSize = 2. Dequeue 15, add to level list, no children. Dequeue 7, add to level list, no children. Level list = [15, 7]. Result = [[3], [9, 20], [15, 7]]. Queue is empty — done.

The three edge cases to handle are straightforward. An empty tree (null root) returns an empty list. A single-node tree returns [[root.val]]. A completely skewed tree — where every node has only a left or only a right child — produces one value per level, resulting in a list of single-element lists.

Time complexity is O(n) where n is the number of nodes, since every node is enqueued and dequeued exactly once. Space complexity is O(w) where w is the maximum width of the tree — in the worst case a complete binary tree, this is roughly n/2 nodes at the last level. The result list itself also takes O(n) space.

The skewed tree case is worth noting because it reveals that BFS does not always use more memory than DFS. For a skewed tree with n nodes, the BFS queue never holds more than one node at a time, while DFS recursion uses O(n) stack frames.

LeetCode 102 is not just a standalone problem — it is the BFS tree template. The queue initialization, the level-size loop, the child enqueueing — this exact skeleton appears in every level-based tree problem. Once you have internalized it, variations become straightforward modifications.

For Zigzag Level Order Traversal (#103), you alternate the direction you append values at each level. For Binary Tree Right Side View (#199), you only keep the last node in each level list. For Average of Levels (#637), you compute the mean of each level list instead of returning raw values. The template stays the same.

Practice this problem until the BFS template is automatic. Use YeetCode flashcards to drill the pattern: what data structure, what loop structure, what the size variable tracks. When this template lives in muscle memory, you will recognize level-order variants in seconds during your next coding interview.