EASY NC#46 Blind #62 Trees
226. Invert Binary Tree
📖 Problem
Given the root of a binary tree, invert the tree, and return its root.
🧠 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
- •In-place array updates
- •Sorted array traversal
- •Boundary condition checks
Logical Thinking Concepts
- •Define invariants before coding
- •Check edge cases first (`[]`, single element, duplicates)
- •Estimate time/space before implementation
- •Apply Two Pointers reasoning pattern
- •Apply BFS reasoning pattern
- •Apply Recursion reasoning pattern
💡 Approach
- → Swap left and right children recursively
- → Base case: null node returns null
- → Post-order traversal (process children before current node)
- → Time: O(n), Space: O(h) where h is tree height
🛠️ Hints & Pitfalls
Hints
- •Swap left and right children recursively
- •Base case: null node returns null
- •Post-order traversal (process children before current node)
Common Pitfalls
- •Time: O(n), Space: O(h) where h is tree height
🧪 Test Cases
Hidden tests on submit: 3
Test Case 1
Not run Input:
invertTree(buildTree([4,2,7,1,3,6,9])); Expected:
{"val":4,"left":{"val":7,"left":{"val":9,"left":null,"right":null},"right":{"val":6,"left":null,"right":null}},"right":{"val":2,"left":{"val":3,"left":null,"right":null},"right":{"val":1,"left":null,"right":null}}} Test Case 2
Not run Input:
invertTree(buildTree([2,1,3])); Expected:
{"val":2,"left":{"val":3,"left":null,"right":null},"right":{"val":1,"left":null,"right":null}} Test Case 3
Not run Input:
invertTree(buildTree([])); Expected:
null 📝 Code Editor
📤 Output