Zigzag Level Order Traversal in DSA Python - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is Zigzag Level Order Traversal in a binary tree?

It is a way to visit nodes level by level, but alternating the direction of traversal at each level. For example, left to right on one level, then right to left on the next.

beginner

Why do we use a flag or boolean variable in Zigzag Level Order Traversal?

To keep track of the current direction of traversal (left to right or right to left) and switch it after completing each level.

intermediate

Which data structure is commonly used to implement Zigzag Level Order Traversal?

A queue is used to process nodes level by level, and sometimes a stack or reversing the list is used to handle the zigzag order.

intermediate

How does reversing the list of nodes at a level help in Zigzag traversal?

When the direction is right to left, reversing the collected nodes of that level before adding to the result gives the correct order.

beginner

What is the time complexity of Zigzag Level Order Traversal?

O(n), where n is the number of nodes in the tree, because each node is visited exactly once.

In Zigzag Level Order Traversal, what changes at each level?

A The direction of traversal

B The number of nodes

C The tree structure

D The node values

The traversal direction alternates between left to right and right to left at each level.

Which data structure is primarily used to hold nodes while performing level order traversal?

A Stack

B Queue

C Linked List

D Hash Map

A queue is used to process nodes level by level in order.

How do we handle the right to left traversal in Zigzag Level Order Traversal?

A Reverse the list of nodes at that level

B Change the tree structure

C Use a stack instead of a queue

D Skip the level

Reversing the list of nodes collected at that level gives the right to left order.

What is the space complexity of Zigzag Level Order Traversal in the worst case?

A O(n^2)

B O(log n)

C O(1)

D O(n)

In the worst case, the queue can hold all nodes at the largest level, which can be O(n).

Which of these is NOT a correct step in Zigzag Level Order Traversal?

A Use a queue to process nodes level by level

B Toggle direction after each level

C Reverse the entire tree after each level

D Collect nodes of each level before adding to result

Reversing the entire tree is not part of the traversal; only the order of nodes at each level is reversed if needed.

Explain how Zigzag Level Order Traversal works step-by-step on a binary tree.

Think about how you visit nodes left to right, then right to left, and so on.

Describe the role of data structures in implementing Zigzag Level Order Traversal.

Focus on how nodes are stored and processed during traversal.

Practice

1. What does the Zigzag Level Order Traversal of a binary tree do differently compared to a normal level order traversal?

easy

A. It traverses the tree using depth-first search.

B. It visits nodes only at even levels.

C. It visits nodes in reverse order of their values.

D. It alternates the direction of traversal at each level, left to right then right to left.

Solution

1. Step 1: Understand normal level order traversal

   Normal level order traversal visits nodes level by level from left to right.

2. Step 2: Identify zigzag difference

   Zigzag traversal changes direction each level: first left to right, then right to left, and so on.

3. Final Answer:

   It alternates the direction of traversal at each level, left to right then right to left. -> Option D

4. Quick Check:

   Zigzag = alternate directions [OK]

Hint: Remember zigzag means direction flips each level [OK]

Common Mistakes:

• Thinking it only visits even levels
• Confusing with reverse value order
• Mixing with depth-first search

2. Which of the following Python code snippets correctly initializes a queue for level order traversal in zigzag pattern?

easy

A. from collections import deque
queue = deque([root])

B. queue = list(root)

C. queue = set(root)

D. queue = root.pop()

Solution

1. Step 1: Identify correct queue initialization

   For level order traversal, a queue is needed; deque from collections is ideal.

2. Step 2: Check each option

   from collections import deque queue = deque([root]) uses deque with root node inside, correct for BFS. Others misuse list, set, or pop method.

3. Final Answer:

   from collections import deque\nqueue = deque([root]) -> Option A

4. Quick Check:

   Use deque for queue in BFS [OK]

Hint: Use deque with root inside for BFS queue [OK]

Common Mistakes:

• Using list instead of deque for queue
• Using set which is unordered
• Calling pop on root which is a node, not a list

3. What is the output of the zigzag level order traversal for the binary tree below?

    1
   / \
  2   3
 / \   \
4   5   6

Assuming the traversal starts left to right at level 0.

medium

A. [[1], [2, 3], [6, 5, 4]]

B. [[1], [3, 2], [6, 5, 4]]

C. [[1], [3, 2], [4, 5, 6]]

D. [[1], [2, 3], [4, 5, 6]]

Solution

1. Step 1: Traverse level 0 left to right

   Level 0 has node 1, so output is [1].

2. Step 2: Traverse level 1 right to left

   Level 1 nodes are 2 and 3; reversed order is [3, 2].

3. Step 3: Traverse level 2 left to right

   Level 2 nodes are 4, 5, 6; order is [4, 5, 6].

4. Final Answer:

   [[1], [3, 2], [4, 5, 6]] -> Option C

5. Quick Check:

   Zigzag levels alternate direction [OK]

Hint: Alternate direction each level starting left to right [OK]

Common Mistakes:

• Not reversing the order on odd levels
• Reversing the wrong levels
• Mixing node values order

4. The following code attempts zigzag level order traversal but produces incorrect output. What is the main error?

def zigzag(root):
    if not root:
        return []
    from collections import deque
    queue = deque([root])
    result = []
    left_to_right = True
    while queue:
        level_size = len(queue)
        level_nodes = []
        for _ in range(level_size):
            node = queue.popleft()
            if left_to_right:
                level_nodes.append(node.val)
            else:
                level_nodes.insert(0, node.val)
            if node.left:
                queue.append(node.left)
            if node.right:
                queue.append(node.right)
        result.append(level_nodes)
        # Missing toggle of direction here
    return result

medium

A. The direction flag 'left_to_right' is never toggled inside the loop.

B. The queue is not initialized correctly.

C. Nodes are appended to the wrong end of the queue.

D. The function does not check if root is null.

Solution

1. Step 1: Analyze direction toggle usage

   The variable 'left_to_right' controls order but is never changed inside the loop.

2. Step 2: Understand effect of missing toggle

   Without toggling, traversal direction stays same every level, so no zigzag effect.

3. Final Answer:

   The direction flag 'left_to_right' is never toggled inside the loop. -> Option A

4. Quick Check:

   Toggle direction each level [OK]

Hint: Toggle direction flag after each level [OK]

Common Mistakes:

• Forgetting to toggle direction
• Incorrect queue initialization
• Misusing queue append/popleft

5. Given a binary tree, how would you modify the zigzag level order traversal to return the sum of node values at each level instead of the nodes themselves?

hard

A. Reverse the entire result list at the end.

B. Accumulate the sum of values in 'level_nodes' before appending to result.

C. Use a stack instead of a queue to store nodes.

D. Append node values directly to result without using 'level_nodes'.

Solution

1. Step 1: Understand current storage of nodes per level

   Currently, nodes are collected in 'level_nodes' list for each level.

2. Step 2: Modify to sum values

   Instead of appending the list, calculate sum(level_nodes) and append that sum to result.

3. Final Answer:

   Accumulate the sum of values in 'level_nodes' before appending to result. -> Option B

4. Quick Check:

   Sum level nodes before appending [OK]

Hint: Sum values in level list before adding to result [OK]

Common Mistakes:

• Appending nodes instead of sums
• Using stack instead of queue incorrectly
• Reversing result instead of levels