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

Time Complexity: Tree Traversal Level Order BFS

O(n)

Understanding Time Complexity

We want to understand how the time needed to visit all nodes in a tree grows as the tree gets bigger.

Specifically, how does the level order traversal (BFS) behave when the tree size increases?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

public void LevelOrderTraversal(TreeNode root) {
    if (root == null) return;
    Queue queue = new Queue();
    queue.Enqueue(root);
    while (queue.Count > 0) {
        TreeNode current = queue.Dequeue();
        Console.Write(current.val + " ");
        if (current.left != null) queue.Enqueue(current.left);
        if (current.right != null) queue.Enqueue(current.right);
    }
}
    

This code visits every node in the tree level by level, from top to bottom and left to right.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Visiting each node once by dequeuing from the queue.
• How many times: Exactly once for each node in the tree.

How Execution Grows With Input

As the number of nodes in the tree grows, the number of visits grows at the same rate.

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

Pattern observation: The work grows directly in proportion to the number of nodes.

Final Time Complexity

Time Complexity: O(n)

This means the time to complete the traversal grows linearly with the number of nodes in the tree.

Common Mistake

[X] Wrong: "Since the tree has multiple levels, the traversal must be slower than linear, maybe quadratic."

[OK] Correct: Each node is visited exactly once, so the total work is proportional to the number of nodes, not the number of levels squared.

Interview Connect

Understanding level order traversal time helps you explain how tree algorithms scale, a key skill in many coding interviews.

Self-Check

"What if we changed the tree to a graph with cycles? How would the time complexity change if we don't track visited nodes?"

Practice

1. What does Level Order Traversal (BFS) of a binary tree do?

easy

A. Visits nodes level by level from top to bottom, left to right

B. Visits nodes in preorder (root, left, right)

C. Visits nodes in inorder (left, root, right)

D. Visits nodes in postorder (left, right, root)

Solution

1. Step 1: Understand Level Order Traversal

   Level order traversal visits nodes starting from the root, then all nodes at the next level, and so on, moving left to right at each level.

2. Step 2: Compare with other traversals

   Preorder, inorder, and postorder are depth-first traversals, not level by level.

3. Final Answer:

   Visits nodes level by level from top to bottom, left to right -> Option A

4. Quick Check:

   Level order = BFS = top to bottom, left to right [OK]

Hint: Level order means visit nodes by levels, not depth [OK]

Common Mistakes:

• Confusing level order with preorder traversal
• Thinking level order is depth-first
• Mixing up left-to-right with right-to-left order

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

easy

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

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

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

D. Dictionary queue = new Dictionary();

Solution

1. Step 1: Identify the correct data structure for BFS

   BFS uses a queue to process nodes in FIFO order.

2. Step 2: Check the code snippet

   Queue<TreeNode> queue = new Queue<TreeNode>(); uses Queue<TreeNode>, which is correct. Others use List, Stack, or Dictionary, which are incorrect for BFS queue.

3. Final Answer:

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

4. Quick Check:

   Queue for BFS = Queue<TreeNode> [OK]

Hint: BFS always uses a queue, not stack or list [OK]

Common Mistakes:

• Using Stack instead of Queue for BFS
• Using List which lacks FIFO behavior
• Confusing Dictionary with queue

3. What is the output of the following C# code performing level order traversal on this tree?

    1
   / \
  2   3
 /     \
4       5

// Code snippet:
Queue<TreeNode> queue = new Queue<TreeNode>();
queue.Enqueue(root);
while(queue.Count > 0) {
  TreeNode node = queue.Dequeue();
  Console.Write(node.val + " ");
  if(node.left != null) queue.Enqueue(node.left);
  if(node.right != null) queue.Enqueue(node.right);
}

medium

A. 4 2 5 3 1

B. 1 2 3 4 5

C. 1 3 2 5 4

D. 2 4 3 5 1

Solution

1. Step 1: Trace the queue operations

   Start with root (1). Dequeue 1, enqueue 2 and 3. Output: 1 Dequeue 2, enqueue 4. Output: 1 2 Dequeue 3, enqueue 5. Output: 1 2 3 Dequeue 4, no children. Output: 1 2 3 4 Dequeue 5, no children. Output: 1 2 3 4 5

2. Step 2: Confirm output order

   The output matches level order: 1 2 3 4 5

3. Final Answer:

   1 2 3 4 5 -> Option B

4. Quick Check:

   Level order output = 1 2 3 4 5 [OK]

Hint: Follow queue enqueue/dequeue step-by-step [OK]

Common Mistakes:

• Mixing preorder with level order output
• Forgetting to enqueue children in correct order
• Printing nodes before dequeue

4. Identify the bug in this C# code snippet for level order traversal:

Queue<TreeNode> queue = new Queue<TreeNode>();
queue.Enqueue(root);
while(queue.Count > 0) {
  TreeNode node = queue.Dequeue();
  Console.Write(node.val + " ");
  if(node.left == null) queue.Enqueue(node.left);
  if(node.right == null) queue.Enqueue(node.right);
}

medium

A. It prints node values after enqueueing children

B. It enqueues null children causing runtime errors

C. It uses Stack instead of Queue

D. It never enqueues any children

Solution

1. Step 1: Analyze the if conditions for enqueue

   The code enqueues children only if node.left == null and node.right == null, which means it enqueues children only if they are null (no child).

2. Step 2: Understand the effect

   Since children are null, it never enqueues actual child nodes, so traversal stops after root.

3. Final Answer:

   It never enqueues any children -> Option D

4. Quick Check:

   Check conditions: should be != null, not == null [OK]

Hint: Enqueue children only if they are NOT null [OK]

Common Mistakes:

• Using == null instead of != null in conditions
• Enqueuing null nodes causing errors
• Confusing order of printing and enqueueing

5. Given a binary tree, how can you modify the level order traversal to return a list of lists, where each inner list contains nodes at that level?

hard

A. Traverse the tree inorder and group nodes by their values

B. Use recursion to visit nodes and store them by depth

C. Use a queue and track the number of nodes at each level, enqueue children accordingly

D. Use a stack to store nodes and pop them level by level

Solution

1. Step 1: Understand level grouping in BFS

   To group nodes by level, we process nodes level by level. We track how many nodes are at the current level before processing them.

2. Step 2: Implement queue with level size tracking

   At each iteration, get the queue size (number of nodes at current level), dequeue exactly that many nodes, enqueue their children, and store the dequeued nodes in a list for that level.

3. Final Answer:

   Use a queue and track the number of nodes at each level, enqueue children accordingly -> Option C

4. Quick Check:

   Level order + level size tracking = grouped levels [OK]

Hint: Track queue size each loop to separate levels [OK]

Common Mistakes:

• Using recursion without tracking depth
• Using stack which is LIFO, not suitable for level order
• Grouping nodes by value instead of level