Tree Traversal Inorder Left Root Right in DSA Python - Build from Scratch

Tree Traversal Inorder Left Root Right

📖 Scenario: Imagine you have a family tree stored as a simple tree structure. You want to visit each family member in a special order: first visit the left child, then the parent, then the right child. This is called inorder traversal.

🎯 Goal: Build a small binary tree and write code to print the values of the nodes in inorder (left, root, right) order.

📋 What You'll Learn

Create a simple binary tree with three nodes: root, left child, and right child

Add a helper variable to hold the result of the traversal

Write a recursive function to perform inorder traversal (left, root, right)

Print the list of node values after traversal

💡 Why This Matters

🌍 Real World

Tree traversal is used in many applications like searching family trees, organizing files, and parsing expressions.

💼 Career

Understanding tree traversal is important for software developers working with data structures, databases, and algorithms.

1

Create a simple binary tree

Create a class called Node with an __init__ method that takes value, left, and right parameters. Then create three nodes: root with value 1, root.left with value 2, and root.right with value 3.

DSA Python

# Define the Node class and create root, root.left, root.right nodes
# Your code here

Hint

Define a class with __init__ to store value and left/right children. Then create three nodes as described.

2

Create a list to hold traversal results

Create an empty list called result to store the values of nodes visited during inorder traversal.

DSA Python

class Node:
    def __init__(self, value, left=None, right=None):
        self.value = value
        self.left = left
        self.right = right

root = Node(1)
root.left = Node(2)
root.right = Node(3)

# Create an empty list called result
# Your code here

Hint

Just create an empty list named result to store traversal output.

3

Write the inorder traversal function

Write a recursive function called inorder that takes a node parameter. If the node is not None, first call inorder on node.left, then append node.value to result, then call inorder on node.right. Finally, call inorder(root) to start traversal.

DSA Python

class Node:
    def __init__(self, value, left=None, right=None):
        self.value = value
        self.left = left
        self.right = right

root = Node(1)
root.left = Node(2)
root.right = Node(3)

result = []

# Define the inorder function and call it with root
# Your code here

Hint

Use recursion to visit left child, then current node, then right child. Append values to result.

4

Print the inorder traversal result

Write a print statement to display the result list after inorder traversal.

DSA Python

class Node:
    def __init__(self, value, left=None, right=None):
        self.value = value
        self.left = left
        self.right = right

root = Node(1)
root.left = Node(2)
root.right = Node(3)

result = []

def inorder(node):
    if node is not None:
        inorder(node.left)
        result.append(node.value)
        inorder(node.right)

inorder(root)

# Print the result list
# Your code here

Hint

Print the result list to see the inorder traversal output.

Practice

1. What is the order of nodes visited in an inorder traversal of a binary tree?

easy

A. Root, Left subtree, Right subtree

B. Root, Right subtree, Left subtree

C. Right subtree, Root, Left subtree

D. Left subtree, Root, Right subtree

Solution

1. Step 1: Understand inorder traversal definition

   Inorder traversal visits nodes in the order: left subtree first, then root node, then right subtree.

2. Step 2: Apply the order to the tree nodes

   This means we first explore all nodes in the left subtree, then the root, then the right subtree.

3. Final Answer:

   Left subtree, Root, Right subtree -> Option D

4. Quick Check:

   Inorder = Left, Root, Right [OK]

Hint: Remember: Inorder means Left, Root, then Right [OK]

Common Mistakes:

• Confusing inorder with preorder or postorder
• Visiting root before left subtree
• Visiting right subtree before root

2. Which of the following Python function definitions correctly implements inorder traversal of a binary tree node root?

easy

A. def inorder(root):
if root:
inorder(root.left)
print(root.val)
inorder(root.right)

B. def inorder(root):
if root:
print(root.val)
inorder(root.left)
inorder(root.right)

C. def inorder(root):
if root:
inorder(root.right)
print(root.val)
inorder(root.left)

D. def inorder(root):
if root:
inorder(root.right)
inorder(root.left)
print(root.val)

Solution

1. Step 1: Recall inorder traversal order

   Inorder visits left subtree, then root, then right subtree.

2. Step 2: Match code to order

   def inorder(root): if root: inorder(root.left) print(root.val) inorder(root.right) calls inorder on left, prints root value, then calls inorder on right, matching inorder traversal.

3. Final Answer:

   Option A code correctly implements inorder traversal -> Option A

4. Quick Check:

   Left, Root, Right order in code [OK]

Hint: Inorder: recursive left call, print root, recursive right call [OK]

Common Mistakes:

• Printing root before left subtree call
• Swapping left and right subtree calls
• Missing base case check for None

3. Given the binary tree below, what is the output of inorder traversal?

    2
   / \
  1   3

class Node: def __init__(self, val): self.val = val self.left = None self.right = None def inorder(root): if root: inorder(root.left) print(root.val, end=' ') inorder(root.right) root = Node(2) root.left = Node(1) root.right = Node(3) inorder(root)

medium

A. 1 2 3

B. 2 1 3

C. 3 2 1

D. 1 3 2

Solution

1. Step 1: Trace inorder traversal on given tree

   Start at root (2), first visit left subtree (1), print 1, then print root 2, then visit right subtree (3), print 3.

2. Step 2: Write output sequence

   The printed sequence is: 1 2 3.

3. Final Answer:

   1 2 3 -> Option A

4. Quick Check:

   Inorder output = 1 2 3 [OK]

Hint: Inorder prints nodes in ascending order for BST [OK]

Common Mistakes:

• Printing root before left subtree
• Swapping left and right subtree outputs
• Confusing preorder or postorder output

4. The following code is intended to perform inorder traversal but produces no output. What is the error?

class Node:
    def __init__(self, val):
        self.val = val
        self.left = None
        self.right = None

def inorder(root):
    inorder(root.left)
    print(root.val, end=' ')
    inorder(root.right)

root = Node(1)
inorder(root)

medium

A. The root node is never created

B. The print statement is outside the function

C. The base case is missing; should check if root is None first

D. The function calls right subtree before printing root

Solution

1. Step 1: Analyze base case in inorder function

   The function checks root.left and root.right but never checks if root itself is None, which causes errors or no output if root is None.

2. Step 2: Identify missing base case effect

   Without checking if root is None, the function may fail or skip printing when called with None, so adding if root is None: return fixes it.

3. Final Answer:

   The base case is missing; should check if root is None first -> Option C

4. Quick Check:

   Check root None before recursion [OK]

Hint: Always check if node is None before recursion [OK]

Common Mistakes:

• Not checking if root is None before accessing attributes
• Incorrect order of print and recursive calls
• Forgetting to create root node before traversal

5. You have a binary tree where some nodes have only one child. Which inorder traversal output correctly represents the tree below?

      5
     / \
    3   8
     \    \
      4    9

Options: A) 3 5 4 8 9 B) 4 3 5 9 8 C) 3 4 5 8 9 D) 5 3 4 8 9

hard

A. 3 5 4 8 9

B. 3 4 5 8 9

C. 4 3 5 9 8

D. 5 3 4 8 9

Solution

1. Step 1: Perform inorder traversal on given tree

   Start at root (5), visit left subtree: node 3 has no left child, so print 3, then visit right child 4, print 4. Then print root 5. Then visit right subtree: node 8 has no left child, print 8, then right child 9, print 9.

2. Step 2: Write the output sequence

   The output is: 3 4 5 8 9.

3. Final Answer:

   3 4 5 8 9 -> Option B

4. Quick Check:

   Inorder visits left, root, right even if child missing [OK]

Hint: Inorder visits left subtree fully before root, even if missing nodes [OK]

Common Mistakes:

• Skipping nodes with only one child
• Printing root before left subtree
• Confusing preorder or postorder output