Zigzag Level Order Traversal in DSA C# - Time & Space Complexity

Time Complexity: Zigzag Level Order Traversal

O(n)

Understanding Time Complexity

We want to understand how the time needed to traverse a tree in zigzag order changes as the tree grows.

Specifically, how does the number of nodes affect the work done?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

This code traverses a binary tree level by level, alternating the order of values between left-to-right and right-to-left.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Visiting each node once in the while loop and inner for loop.
• How many times: Each node is processed exactly once, so n times where n is the number of nodes.

How Execution Grows With Input

As the number of nodes grows, the code visits each node once, so the work grows directly with n.

Input Size (n)	Approx. Operations
10	About 10 node visits
100	About 100 node visits
1000	About 1000 node visits

Pattern observation: The operations increase linearly as the tree size increases.

Final Time Complexity

Time Complexity: O(n)

This means the time to complete the traversal grows in direct proportion to the number of nodes in the tree.

Common Mistake

[X] Wrong: "Reversing the list for every right-to-left level makes the time complexity worse than linear O(n²)."

[OK] Correct: Reversing a list of size k takes O(k) time, but the sum of all level sizes k is n, so total reversing time is O(n).

Interview Connect

Understanding this traversal's time complexity helps you explain how tree traversals scale and shows you can analyze mixed operations in loops.

Self-Check

"What if we used a deque to add elements at both ends instead of reversing the list? How would the time complexity change?"

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