Introduction

The Binary Tree Level Order Traversal problem involves traversing a binary tree level by level.

The traversal order is:

Left to RightLevel by Level

This problem is one of the most important applications of:

 Breadth First Search (BFS)

The task is to:

  • return all node values
    grouped level-wise

This problem helps in understanding:

  • tree traversal
  • queue processing
  • BFS traversal
  • level-order processing

Example

 Input:        3

       / \

      9  20

         / \

        15  7

Output:
[[3],[9,20], [15,7]]

Explanation:
Level 1:
3

Level 2:
9 20

Level 3:
15 7

Constraints

Number of nodes:0 <= n <= 2000
-1000 <= Node.val <= 1000

Approach 1 : Brute Force using Height Traversal

Explanation

The simplest way is:

  1. Find tree height
  2. Traverse every level separately
  3. Collect nodes level-wise

This works but repeatedly traversing levels increases complexity.

Steps

  1. Find tree height.
  2. Traverse each level.
  3. Store nodes of current level.
  4. Repeat for all levels.
  5. Return result.

Dry Run

Tree:        3

       / \

      9  20

         / \

        15  7

Height:
3

Traverse Level 1:
3

Traverse Level 2:
9 20

Traverse Level 3:
15 7

Final Result:
[[3],[9,20],[15,7]]

Brute Force Code

Complexity Analysis

Time Complexity: O(n²)Explanation:
Levels are traversed repeatedly.

Space Complexity: O(h)
Explanation:
Recursive traversal uses stack space.

Approach 2 : BFS Queue Traversal

Explanation

The optimized solution uses:

 Queue Based BFS

Idea:

  • process tree level by level
  • queue stores nodes
    of current traversal

Process:

  1. Add root to queue
  2. Traverse one level at a time
  3. Push children into queue
  4. Store level nodes

This naturally performs level-order traversal.

Steps

  1. Create queue.
  2. Insert root node.
  3. Process current level size.
  4. Store level nodes.
  5. Push left and right children.
  6. Return result.

Dry Run

Initial Queue:[3]
Process Level 1:3
Queue:
[9,20]

Process Level 2:
9 20

Queue:
[15,7]

Process Level 3:
15 7

Final Result:
[[3],[9,20],[15,7]]

BFS Queue Code

Complexity Analysis

Time Complexity: O(n)
Explanation:
Each tree node is visited exactly once during BFS traversal.
Space Complexity: O(n)
Explanation:
Queue and result list can store up to n nodes.

Edge Cases

  1. Empty tree
  2. Single node tree
  3. Left skewed tree
  4. Right skewed tree
  5. Complete binary tree

Why This Problem is Important

Binary Tree Level Order Traversal helps in understanding:

  1. BFS traversal
  2. Queue processing
  3. Tree level traversal
  4. Level-order processing
  5. Graph traversal fundamentals

It is one of the most important BFS interview problems.

Real-World Applications

Level-order traversal concepts are used in:

  1. Hierarchical systems
  2. Organization structures
  3. Network traversal
  4. File system traversal
  5. Multi-level processing systems

Common Mistakes

  1. Incorrect level size handling
  2. Forgetting null checks
  3. Mixing levels together
  4. Incorrect child insertion order

Interview Tips

Interviewers often expect:

  1. BFS traversal explanation
  2. Queue usage understanding
  3. Level-order reasoning

Always explain:

  • why queue maintains BFS order
  • how level size separates levels
  • why children are pushed level-wise

Related Questions

  1. Rotten Oranges
  2. Word Ladder
  3. Open the Lock
  4. Zigzag Level Order Traversal
  5. Right Side View

Final Takeaway

The Binary Tree Level Order Traversal problem is a fundamental BFS traversal problem that teaches queue-based level-order processing techniques. Understanding this problem builds a strong foundation for advanced tree and graph traversal interview problems.