How to Solve Inorder Traversal on LeetCode
The Inorder Traversal problem is a fundamental binary tree challenge frequently asked in coding interviews. It checks your understanding of depth-first search (DFS) and tree data structures.
Problem Statement
Given the root of a binary tree, return the inorder traversal of its nodes' values.
Inorder means: visit the left subtree, then the root, then the right subtree.
Example:
Input: root = [1,null,2,3]
Output: [1,3,2]
Why This Problem Matters for Interviews
This problem:
- Validates your grasp of binary trees and traversal orders
- Tests recursive and iterative thinking
- Sets a foundation for harder tree algorithms
Approaches to Inorder Traversal
1. Recursive DFS
Time Complexity: O(n)
Space Complexity: O(n) for the recursion stack in the worst case
def inorderTraversal(root):
res = []
def dfs(node):
if not node:
return
dfs(node.left)
res.append(node.val)
dfs(node.right)
dfs(root)
return res
2. Iterative with Stack
Time Complexity: O(n)
Space Complexity: O(n)
def inorderTraversal(root):
res, stack = [], []
curr = root
while curr or stack:
while curr:
stack.append(curr)
curr = curr.left
curr = stack.pop()
res.append(curr.val)
curr = curr.right
return res
Key Interview Talking Points
- Difference between inorder, preorder, and postorder traversals
- Why recursion matches tree structure
- Stack simulates recursion in iterative method
- Handling empty or skewed trees
Big O Complexity Recap
| Approach | Time Complexity | Space Complexity |
|---|---|---|
| Recursive | O(n) | O(n) |
| Iterative Stack | O(n) | O(n) |
Pro Interview Tips
- Draw out traversal order on sample trees.
- Compare results to preorder/postorder.
- Practice with test cases