Coding Interview Preparation on Telegram: “🎯 CODING CHALLENGE #6

🎯 CODING CHALLENGE #6 - Binary Tree Level Order Traversal
Difficulty: Medium | Asked at: Amazon, Microsoft, LinkedIn

Given a binary tree, return its values level by level (BFS).


        3
       / \
      9   20
         /  \
        15   7

Output: [[3], [9, 20], [15, 7]]

💡 Hint: BFS naturally processes a tree level by level using a queue. The trick is tracking how many nodes belong to the CURRENT level before you start adding next-level nodes to the same queue.

Solution:

python
from collections import deque

def level_order(root):
    if not root:
        return []

    result = []
    queue = deque([root])

    while queue:
        level_size = len(queue)
        current_level = []

        for _ in range(level_size):
            node = queue.popleft()
            current_level.append(node.val)
            if node.left:
                queue.append(node.left)
            if node.right:
                queue.append(node.right)

        result.append(current_level)

    return result

Complexity: O(n) time and space - every node is visited and queued exactly once.

Common mistake: Forgetting to snapshot level_size = len(queue) BEFORE the inner loop starts. If you check len(queue) inside the loop, it changes as you enqueue children, and your levels get mixed together.

BFS with a queue vs. DFS with recursion - do you know when to reach for each? That's often the actual follow-up question here 👇