MEDIUM NC#53 Blind #64 Trees / BFS
102. Binary Tree Level Order Traversal
📖 Problem
Given the root of a binary tree, return the level order traversal of its nodes' values (i.e., from left to right, level by level).
🧠 Visual Learning Aid
1 Model the input into the right structure
2 Choose the core technique and invariant
3 Execute step-by-step with a sample
4 Validate complexity and edge cases
JS/TS Refreshers
- •Array methods (`push`, `pop`, `shift`, `slice`)
- •Object/Map/Set usage patterns
- •Function parameter and return typing
- •Level-order traversal
- •Queue discipline
- •Shortest-step interpretation
Logical Thinking Concepts
- •Define invariants before coding
- •Check edge cases first (`[]`, single element, duplicates)
- •Estimate time/space before implementation
- •Apply BFS reasoning pattern
💡 Approach
- → Use BFS (queue) for level order traversal
- → Process all nodes at current level before moving to next
- → Collect each level's values separately
- → Time: O(n), Space: O(w) where w is max width
🛠️ Hints & Pitfalls
Hints
- •Use BFS (queue) for level order traversal
- •Process all nodes at current level before moving to next
- •Collect each level's values separately
Common Pitfalls
- •Time: O(n), Space: O(w) where w is max width
🧪 Test Cases
Hidden tests on submit: 1
Test Case 1
Not run Input:
levelOrder(buildTree([3,9,20,null,null,15,7])); Expected:
[[3], [9,20], [15,7]] Test Case 2
Not run Input:
levelOrder(buildTree([1])); Expected:
[[1]] Test Case 3
Not run Input:
levelOrder(buildTree([])); Expected:
[] 📝 Code Editor
📤 Output