In order Traversal

Unit 8 Session 1 (Click for link to problem statements)

Problem Highlights

• 💡 Difficulty: Easy
• ⏰ Time to complete: 10 mins
• 🛠️ Topics: Trees, Binary Trees, Tree Traversal

1: U-nderstand

Understand what the interviewer is asking for by using test cases and questions about the problem.

• Established a set (2-3) of test cases to verify their own solution later.
• Established a set (1-2) of edge cases to verify their solution handles complexities.
• Have fully understood the problem and have no clarifying questions.
• Have you verified any Time/Space Constraints for this problem?

• Question: What is inorder traversal?
  • Answer: In an inorder traversal of a binary tree, each node is visited in the following order: left child, node itself, then right child.

HAPPY CASE
Input: TreeNode(1, TreeNode(2), TreeNode(3))
Output: [2, 1, 3]
Explanation: Inorder traversal visits the left child first, then the node, and finally the right child.

EDGE CASE
Input: TreeNode(1, None, TreeNode(2))
Output: [1, 2]
Explanation: With no left child, the root is visited first, then the right child.

2: M-atch

Match what this problem looks like to known categories of problems, e.g. Linked List or Dynamic Programming, and strategies or patterns in those categories.

This problem is a standard tree traversal problem, specifically an inorder traversal which is fundamental in binary tree operations.

3: P-lan

Plan the solution with appropriate visualizations and pseudocode.

General Idea: Recursively traverse the tree to access each node in the inorder sequence.

1) Recursively visit the left subtree.
2) Visit the current node (append its value).
3) Recursively visit the right subtree.

⚠️ Common Mistakes

• Not maintaining the order of traversal which could lead to incorrect results.

4: I-mplement

Implement the code to solve the algorithm.

def inorder_helper(current_node, values):
    if not current_node:
        return values
    inorder_helper(current_node.left, values)  # Traverse the left subtree
    values.append(current_node.val)  # Visit the node
    inorder_helper(current_node.right, values)  # Traverse the right subtree
    return values

def inorder_traversal(root):
    values = []
    return inorder_helper(root, values)

5: R-eview

Review the code by running specific example(s) and recording values (watchlist) of your code's variables along the way.

• Step through the code with a test tree to ensure that the inorder sequence is correctly assembled.

6: E-valuate

Evaluate the performance of your algorithm and state any strong/weak or future potential work.

• Time Complexity: O(n) where n is the number of nodes in the tree. Each node is visited exactly once.
• Space Complexity: O(n) for the recursion stack in the worst case when the tree is skewed.