Zigzag Level Order Traversal in DSA C# - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Zigzag Level Order Traversal on a small tree

What is the output of the following C# code performing zigzag level order traversal on the given binary tree?

DSA C#

public class TreeNode {
    public int val;
    public TreeNode left;
    public TreeNode right;
    public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
        this.val = val;
        this.left = left;
        this.right = right;
    }
}

public IList<IList<int>> ZigzagLevelOrder(TreeNode root) {
    var results = new List<IList<int>>();
    if (root == null) return results;
    var currentLevel = new List<TreeNode> { root };
    bool leftToRight = true;
    while (currentLevel.Count > 0) {
        var levelValues = new List<int>();
        var nextLevel = new List<TreeNode>();
        foreach (var node in currentLevel) {
            levelValues.Add(node.val);
            if (node.left != null) nextLevel.Add(node.left);
            if (node.right != null) nextLevel.Add(node.right);
        }
        if (!leftToRight) levelValues.Reverse();
        results.Add(levelValues);
        currentLevel = nextLevel;
        leftToRight = !leftToRight;
    }
    return results;
}

// Tree structure:
//       1
//      / \
//     2   3
//    / \   \
//   4   5   6

var root = new TreeNode(1,
    new TreeNode(2, new TreeNode(4), new TreeNode(5)),
    new TreeNode(3, null, new TreeNode(6))
);
var result = ZigzagLevelOrder(root);
foreach(var level in result) {
    Console.WriteLine(string.Join(",", level));
}

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

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

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

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

💡 Hint

Remember that the traversal direction alternates at each level.

The traversal starts left to right at level 1: [1].

Next level is right to left: [3, 2].

Next level is left to right again: [4, 5, 6].

❓ Predict Output

intermediate

Output for Zigzag Traversal on single node tree

What is the output of zigzag level order traversal on a tree with only one node (value 10)?

DSA C#

var root = new TreeNode(10);
var result = ZigzagLevelOrder(root);
foreach(var level in result) {
    Console.WriteLine(string.Join(",", level));
}

A [[10]]

B []

C [[]]

D [[0]]

💡 Hint

A single node tree has only one level.

The traversal returns one level with the single node value 10.

🔧 Debug

advanced

Identify the error in zigzag traversal code

What error will this code produce when run on a non-empty tree?

DSA C#

public IList<IList<int>> ZigzagLevelOrder(TreeNode root) {
    var results = new List<IList<int>>();
    if (root == null) return results;
    var queue = new Queue<TreeNode>();
    queue.Enqueue(root);
    bool leftToRight = true;
    while (queue.Count > 0) {
        int size = queue.Count;
        var levelValues = new List<int>();
        for (int i = 0; i < size; i++) {
            var node = queue.Dequeue();
            if (leftToRight) {
                levelValues.Add(node.val);
            } else {
                levelValues.Insert(0, node.val);
            }
            if (node.left != null) queue.Enqueue(node.left);
            if (node.right != null) queue.Enqueue(node.right);
        }
        results.Add(levelValues);
        leftToRight = !leftToRight;
    }
    return results;
}

// Usage with a tree root

A NullReferenceException when accessing node.left or node.right

B No error; code runs correctly and returns zigzag traversal

C InvalidOperationException when calling queue.Dequeue on empty queue

D StackOverflowException due to infinite recursion

💡 Hint

Check how the queue is managed and how nodes are enqueued and dequeued.

The code correctly uses a queue and alternates insertion order to achieve zigzag traversal without errors.

🚀 Application

advanced

Number of levels in zigzag traversal output

Given a complete binary tree with 7 nodes, how many levels will the zigzag level order traversal output contain?

A 2

B 7

C 4

D 3

💡 Hint

Count the height of a complete binary tree with 7 nodes.

A complete binary tree with 7 nodes has height 3, so traversal output has 3 levels.

🧠 Conceptual

expert

Why use a deque or list reversal in zigzag traversal?

In zigzag level order traversal, why do we reverse the list of node values or use a deque to insert at front on alternate levels?

A To remove duplicate node values from the traversal

B To sort the node values in ascending order at each level

C To alternate the order of node values between left-to-right and right-to-left at each level

D To ensure all nodes are visited only once

💡 Hint

Think about the zigzag pattern and how order changes per level.

Reversing or inserting at front changes the order of nodes to achieve the zigzag pattern alternating directions.

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

B. It visits nodes only on the left subtree.

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

D. It traverses nodes randomly without any order.

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 alternates 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 A

4. Quick Check:

   Zigzag = alternate directions [OK]

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

Common Mistakes:

• Thinking it only visits left subtree
• Assuming traversal is always left to right
• Confusing zigzag with reverse order

2. Which of the following C# code snippets correctly initializes a queue to perform a zigzag level order traversal?

easy

A. List<TreeNode> queue = new List<TreeNode>();

B. Queue<TreeNode> queue = new Queue<TreeNode>();

C. Stack<TreeNode> queue = new Stack<TreeNode>();

D. Dictionary queue = new Dictionary();

Solution

1. Step 1: Identify data structure for BFS

   Queue is used for breadth-first search to process nodes level by level.

2. Step 2: Check code snippet correctness

   Queue<TreeNode> queue = new Queue<TreeNode>(); correctly declares a Queue of TreeNode objects, suitable for traversal.

3. Final Answer:

   Queue<TreeNode> queue = new Queue<TreeNode>(); -> Option B

4. Quick Check:

   Queue for BFS = Queue<TreeNode> queue = new Queue<TreeNode>(); [OK]

Hint: Use Queue for level order traversal in C# [OK]

Common Mistakes:

• Using Stack instead of Queue for BFS
• Using List which lacks FIFO behavior
• Using Dictionary which is not suitable here

3. Given the binary tree:

    1
   / \
  2   3
 / \   \
4   5   6

What is the output of zigzag level order traversal?

medium

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

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

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

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

Solution

1. Step 1: Traverse level 0 left to right

   Level 0 nodes: [1]

2. Step 2: Traverse level 1 right to left

   Level 1 nodes: [2, 3], reversed to [3, 2]

3. Step 3: Traverse level 2 left to right

   Level 2 nodes: [4, 5, 6]

4. Final Answer:

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

5. Quick Check:

   Zigzag levels = [[1], [3, 2], [4, 5, 6]] [OK]

Hint: Alternate direction each level, reverse nodes on odd levels [OK]

Common Mistakes:

• Not reversing nodes on alternate levels
• Mixing node order in last level
• Confusing left and right child order

4. Consider this C# code snippet for zigzag traversal:

if (level % 2 == 0)
    result.Add(currentLevel);
else
    result.Add(currentLevel.Reverse().ToList());

What is the bug in this code?

medium

A. It reverses the list on even levels instead of odd levels.

B. It reverses the list on odd levels instead of even levels.

C. It reverses the original list causing side effects.

D. It does not reverse the list at all.

Solution

1. Step 1: Understand zigzag pattern

   Zigzag reverses nodes on odd levels (1, 3, 5...), not even levels (0, 2, 4...).

2. Step 2: Analyze code condition

   The condition correctly identifies even/odd levels for reversal.

3. Step 3: Identify the actual issue

   currentLevel.Reverse() (List<TreeNode>.Reverse) modifies the list in place and returns void, causing side effects and a compilation error on .ToList().

4. Final Answer:

   It reverses the original list causing side effects. -> Option C

5. Quick Check:

   List.Reverse() in-place modifies original [OK]

Hint: Reverse nodes only on odd levels (1, 3, 5...) [OK]

Common Mistakes:

• Reversing on even levels instead of odd
• Modifying original list causing bugs
• Not reversing at all

5. You want to perform zigzag level order traversal on a binary tree but only collect nodes with even values. Which modification is best to apply?

hard

A. Use a stack instead of a queue to collect only even nodes.

B. Reverse the entire result list after traversal to get even nodes first.

C. Skip reversing on odd levels to keep order consistent for even nodes.

D. Filter nodes by even value before adding them to the current level list during traversal.

Solution

1. Step 1: Identify filtering requirement

   We want only even-valued nodes in the output, so filter nodes during traversal.

2. Step 2: Apply filtering during level collection

   Filter nodes by checking if value is even before adding to current level list.

3. Step 3: Keep zigzag logic intact

   After filtering, apply zigzag reversal as usual on filtered nodes.

4. Final Answer:

   Filter nodes by even value before adding them to the current level list during traversal. -> Option D

5. Quick Check:

   Filter during traversal = Filter nodes by even value before adding them to the current level list during traversal. [OK]

Hint: Filter nodes while collecting each level, then zigzag [OK]

Common Mistakes:

• Reversing entire result instead of filtering
• Skipping reversal breaks zigzag pattern
• Using stack incorrectly for filtering