MEDIUM NC#58 Blind #67 Trees
105. Construct Binary Tree from Preorder and Inorder Traversal
📖 Problem
Given two integer arrays preorder and inorder where preorder is the preorder traversal of a binary tree and inorder is the inorder traversal of the same tree, construct and return the binary tree.
🧠 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 Recursion reasoning pattern
💡 Approach
- → First element in preorder is always the root
- → Find root in inorder - elements left are left subtree
- → Recursively construct left and right subtrees
- → Use hash map for O(1) root lookups in inorder
- → Time: O(n), Space: O(n)
🧭 Prerequisites
🛠️ Hints & Pitfalls
Hints
- •First element in preorder is always the root
- •Find root in inorder - elements left are left subtree
- •Recursively construct left and right subtrees
Common Pitfalls
- •Use hash map for O(1) root lookups in inorder
- •Time: O(n), Space: O(n)
🧪 Test Cases
Hidden tests on submit: 1
Test Case 1
Not run Input:
arrayToTree(left: number, right: number); Expected:
Computed from hidden reference Test Case 2
Not run Input:
arrayToTree(left, inorderMap.get(rootValue)! - 1); Expected:
Computed from hidden reference Test Case 3
Not run Input:
arrayToTree(inorderMap.get(rootValue)! + 1, right); Expected:
Computed from hidden reference 📝 Code Editor
📤 Output