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

Recall & Review

beginner

What is Level Order Traversal in a tree?

Level Order Traversal visits nodes level by level from top to bottom and left to right within each level.

beginner

Which data structure is commonly used to implement Level Order Traversal (BFS) in trees?

A queue is used to keep track of nodes to visit in order, ensuring nodes are processed level by level.

beginner

What is the time complexity of Level Order Traversal on a tree with n nodes?

The time complexity is O(n) because each node is visited exactly once.

intermediate

In Level Order Traversal, what happens when a node is dequeued?

When a node is dequeued, its value is processed (e.g., printed), and its children are enqueued to be processed later.

intermediate

Why is Level Order Traversal also called Breadth-First Search (BFS)?

Because it explores all nodes at the current depth before moving to nodes at the next depth, spreading out like a wave.

Which data structure is essential for implementing Level Order Traversal?

A Stack

B Queue

C Hash Map

D Linked List

A queue helps process nodes in the order they appear level by level.

What is the order of visiting nodes in Level Order Traversal?

A Top to bottom, left to right

B Bottom to top, right to left

C Left subtree, root, right subtree

D Right subtree, root, left subtree

Level Order visits nodes level by level from top to bottom and left to right.

What is the time complexity of Level Order Traversal for a tree with n nodes?

A O(1)

B O(log n)

C O(n^2)

D O(n)

Each node is visited once, so time complexity is O(n).

What happens to the children of a node during Level Order Traversal?

A They are processed immediately

B They are ignored

C They are enqueued for later processing

D They are deleted

Children are added to the queue to be processed after current nodes.

Level Order Traversal is also known as?

A Breadth-First Search

B Depth-First Search

C Inorder Traversal

D Postorder Traversal

Level Order Traversal explores nodes breadth-wise, so it is called Breadth-First Search.

Explain how Level Order Traversal works on a binary tree using a queue.

Think about visiting nodes level by level.

Describe the difference between Level Order Traversal and Depth-First Traversal.

Compare breadth-first and depth-first approaches.

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