Binary Tree Zigzag Level Order Traversal

Binary Tree Zigzag Level Order Traversal

What is this problem about?

Binary Tree Zigzag Level Order Traversal is a variation of the level order traversal. Instead of reading every level from left to right, you alternate the direction for each level. The first level is left-to-right, the second is right-to-left, the third is left-to-right, and so on. This "zigzag" pattern makes for an interesting test of data structure manipulation.

Why is this asked in interviews?

This is a high-signal question for companies like Microsoft, Amazon, and LinkedIn. It tests whether you can adapt the standard BFS pattern and effectively use double-ended data structures like a Deque. It's a great test of attention to detail, specifically regarding when to reverse the output and how to maintain consistency across levels.

Algorithmic pattern used

The primary pattern is Breadth-First Search (BFS). You use a queue to traverse the tree level by level. To handle the zigzagging, you can maintain a boolean flag (e.g., is_left_to_right) that you toggle after each level. When processing a level, if the flag is false, you reverse the list of node values for that level before adding it to the final result.

Example explanation

Consider this tree: 3 / 9 20 /
15 7

1. Level 0: [3] (Left to Right). Result: [[3]].
2. Level 1: [9, 20]. Since it's the second level, we reverse it to [20, 9]. Result: [[3], [20, 9]].
3. Level 2: [15, 7]. Back to Left to Right. Result: [[3], [20, 9], [15, 7]].

Common mistakes candidates make

The most common mistake is overcomplicating the queue logic by trying to push nodes into the queue in different orders for different levels. It’s much simpler to always traverse left-to-right and just reverse the level's list at the end of each odd level. Another mistake is failing to toggle the direction flag correctly.

Interview preparation tip

Practice using a Deque (double-ended queue) for the level results. A deque allows you to add elements to either the front or the back in $O(1)$ time, which can eliminate the need for a separate reversal step and make your code more efficient.