Binary Tree Level Order Traversal – Solution & Complexity

Solution Walkthrough

  1. Step 1: Recognize the Pattern

    The key pattern is: BFS processes one queue level at a time. Identify the state and invariant before coding.

  2. Step 2: Build the Algorithm

    Advance one state transition at a time. Mark or update state before exploring dependent work.

  3. Step 3: Check Edge Cases

    Test empty or minimal input, skewed shapes, duplicates where allowed, and impossible outcomes.

  4. Step 4: Solution and Complexity

    Time: O(n). Space: O(n).

    Java/Go/Rust below are reference solutions for learning the pattern in those languages; only Python and JavaScript can be run and submitted for this problem, since the remote judge doesn't support the TreeNode structure yet.

    def level_order(root: TreeNode) -> list[list[int]]:
        if root is None:
            return []
        answer, queue, head = [], [root], 0
        while head < len(queue):
            level = []
            for _ in range(len(queue) - head):
                node = queue[head]; head += 1
                level.append(node.val)
                if node.left: queue.append(node.left)
                if node.right: queue.append(node.right)
            answer.append(level)
        return answer

FAQ