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

Time Complexity: Tree Traversal Level Order BFS

O(n)

Understanding Time Complexity

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

Specifically, we ask: How does the work change when we use level order traversal (BFS)?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

from collections import deque

def level_order_bfs(root):
    if not root:
        return []
    result = []
    queue = deque([root])
    while queue:
        node = queue.popleft()
        result.append(node.val)
        if node.left:
            queue.append(node.left)
        if node.right:
            queue.append(node.right)
    return result
    

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

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Visiting each node once and adding its children to the queue.
• How many times: Exactly once per node in the tree.

How Execution Grows With Input

As the number of nodes grows, the work grows directly with it.

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

Pattern observation: The number of steps grows in a straight line with the number of nodes.

Final Time Complexity

Time Complexity: O(n)

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

Common Mistake

[X] Wrong: "Level order traversal takes more time because it uses a queue and visits nodes multiple times."

[OK] Correct: Each node is visited exactly once, and the queue just helps keep track of nodes to visit next. So the total work is still proportional to the number of nodes.

Interview Connect

Understanding this traversal's time complexity helps you explain how tree algorithms scale, a key skill in many coding challenges and interviews.

Self-Check

"What if we changed the tree to be very unbalanced, like a linked list? How would the time complexity change?"

Practice

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

easy

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

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

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

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

Solution

1. Step 1: Understand Level Order Traversal

   Level order traversal visits nodes level by level starting from the root, moving left to right on each level.

2. Step 2: Compare with other traversals

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

3. Final Answer:

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

4. Quick Check:

   Level order = BFS = level by level [OK]

Hint: Level order means visiting nodes level by level [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 Python data structure is best suited to implement Level Order Traversal (BFS) of a binary tree?

easy

A. Stack

B. Dictionary

C. Set

D. Queue

Solution

1. Step 1: Identify traversal order needs

   Level order traversal requires visiting nodes in the order they appear level-wise, which is FIFO (first-in, first-out).

2. Step 2: Match data structure to FIFO

   A queue follows FIFO, so it is the best choice. Stack is LIFO, set and dictionary do not maintain order.

3. Final Answer:

   Queue -> Option D

4. Quick Check:

   Level order uses queue (FIFO) [OK]

Hint: Use queue for BFS to process nodes FIFO [OK]

Common Mistakes:

• Using stack instead of queue
• Using set which is unordered
• Using dictionary which is key-value, not order-based

3. What is the output of the following code for level order traversal?

class Node:
    def __init__(self, val):
        self.val = val
        self.left = None
        self.right = None

def level_order(root):
    if not root:
        return []
    result = []
    queue = [root]
    while queue:
        node = queue.pop(0)
        result.append(node.val)
        if node.left:
            queue.append(node.left)
        if node.right:
            queue.append(node.right)
    return result

root = Node(1)
root.left = Node(2)
root.right = Node(3)
root.left.left = Node(4)
root.left.right = Node(5)
print(level_order(root))

medium

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

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

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

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

Solution

1. Step 1: Trace the queue operations

   Start with queue = [1]. Pop 1, add 1 to result, enqueue 2 and 3. Queue = [2,3]. Pop 2, add 2, enqueue 4 and 5. Queue = [3,4,5]. Pop 3, add 3, no children. Queue = [4,5]. Pop 4, add 4. Queue = [5]. Pop 5, add 5. Queue empty.

2. Step 2: Collect the result list

   Result list after traversal is [1, 2, 3, 4, 5].

3. Final Answer:

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

4. Quick Check:

   Level order output matches [1, 2, 3, 4, 5] [OK]

Hint: Follow queue pops and appends step-by-step [OK]

Common Mistakes:

• Mixing left and right child order
• Using stack pop instead of queue pop
• Appending children in wrong order

4. Find the error in this level order traversal code:

def level_order(root):
    if root is None:
        return []
    queue = []
    result = []
    queue.append(root)
    while queue:
        node = queue.pop()
        result.append(node.val)
        if node.left:
            queue.append(node.left)
        if node.right:
            queue.append(node.right)
    return result

medium

A. Using pop() removes last element, should use pop(0) for queue

B. Missing check if root is None

C. Appending children in wrong order

D. Result list is not initialized

Solution

1. Step 1: Identify queue operation mistake

   The code uses queue.pop() which removes the last element (LIFO), but BFS needs FIFO order, so pop(0) should be used.

2. Step 2: Understand impact on traversal order

   Using pop() makes traversal behave like DFS with stack, not BFS, so output will be incorrect.

3. Final Answer:

   Using pop() removes last element, should use pop(0) for queue -> Option A

4. Quick Check:

   pop(0) needed for BFS queue [OK]

Hint: Use pop(0) to dequeue from front in BFS [OK]

Common Mistakes:

• Using pop() instead of pop(0)
• Not initializing result list
• Appending children in wrong order

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

hard

A. Use a queue and track the number of nodes at each level, appending values level-wise

B. Use recursion to visit nodes in preorder and group by depth

C. Use a stack to store nodes and pop them level-wise

D. Traverse the tree inorder and split nodes by value ranges

Solution

1. Step 1: Understand level-wise grouping

   To group nodes by level, we can process nodes level by level using a queue and count how many nodes are at the current level.

2. Step 2: Implement counting and grouping

   At each iteration, record the size of the queue (number of nodes at current level), dequeue exactly that many nodes, collect their values, and enqueue their children for next level.

3. Final Answer:

   Use a queue and track the number of nodes at each level, appending values level-wise -> Option A

4. Quick Check:

   Queue + level count = grouped level order [OK]

Hint: Count nodes per level in queue to group them [OK]

Common Mistakes:

• Using recursion without tracking depth
• Using stack which is LIFO, not suitable
• Trying inorder traversal for level grouping