Tree Traversal Inorder Left Root Right in DSA Python - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the inorder traversal output of this binary tree?

Given the binary tree constructed below, what is the printed inorder traversal (Left, Root, Right)?

DSA Python

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

def inorder(root):
    if root is None:
        return []
    return inorder(root.left) + [root.val] + inorder(root.right)

root = Node(10)
root.left = Node(5)
root.right = Node(15)
root.left.left = Node(3)
root.left.right = Node(7)
root.right.right = Node(18)

result = inorder(root)
print(result)

A [3, 7, 5, 10, 18, 15]

B [10, 5, 3, 7, 15, 18]

C [3, 5, 7, 10, 15, 18]

D [5, 3, 7, 10, 15, 18]

💡 Hint

Remember inorder traversal visits left subtree first, then root, then right subtree.

Inorder traversal visits nodes in the order: left child, root, right child. So for this tree, the order is 3 (leftmost), 5 (parent of 3 and 7), 7, 10 (root), 15, 18.

❓ Predict Output

intermediate

What is the inorder traversal output of this unbalanced tree?

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

DSA Python

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

def inorder(root):
    if root is None:
        return []
    return inorder(root.left) + [root.val] + inorder(root.right)

root = Node(8)
root.left = Node(3)
root.left.right = Node(6)
root.left.right.left = Node(4)
root.left.right.right = Node(7)
root.right = Node(10)
root.right.right = Node(14)
root.right.right.left = Node(13)

result = inorder(root)
print(result)

A [8, 3, 6, 4, 7, 10, 14, 13]

B [3, 4, 6, 7, 8, 10, 13, 14]

C [3, 6, 4, 7, 8, 10, 14, 13]

D [4, 6, 7, 3, 8, 10, 13, 14]

💡 Hint

Follow left subtree fully before root, then right subtree.

Inorder traversal visits left subtree nodes first, then root, then right subtree nodes. The correct order is 3, 4, 6, 7, 8, 10, 13, 14.

🔧 Debug

advanced

What error does this inorder traversal code raise?

Consider the following code for inorder traversal. What error will it raise when run?

DSA Python

def inorder(root):
    if root is None:
        return
    inorder(root.left)
    print(root.val)
    inorder(root.right)

root = None
inorder(root)

A No output, no error

B AttributeError: 'NoneType' object has no attribute 'left'

C TypeError: 'NoneType' object is not callable

D RecursionError: maximum recursion depth exceeded

💡 Hint

Check what happens when root is None at the start.

The function returns immediately when root is None, so no output and no error occurs.

❓ Predict Output

advanced

What is the inorder traversal output of this tree with duplicate values?

Given the binary tree below with duplicate values, what is the inorder traversal output?

DSA Python

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

def inorder(root):
    if root is None:
        return []
    return inorder(root.left) + [root.val] + inorder(root.right)

root = Node(5)
root.left = Node(3)
root.right = Node(5)
root.left.left = Node(3)
root.left.right = Node(4)
root.right.right = Node(7)

result = inorder(root)
print(result)

A [3, 4, 5, 3, 5, 7]

B [3, 4, 3, 5, 7, 5]

C [5, 3, 3, 4, 5, 7]

D [3, 3, 4, 5, 5, 7]

💡 Hint

Duplicates appear in the order they are visited in inorder traversal.

Inorder traversal visits left subtree first, so the two 3s come first, then 4, then root 5, then right 5 and 7.

🧠 Conceptual

expert

How many nodes are visited in inorder traversal of a full binary tree with height 3?

A full binary tree is a tree where every node has 0 or 2 children. If the tree height is 3 (root at height 1), how many nodes will the inorder traversal visit?

A 7

B 8

C 15

D 6

💡 Hint

Number of nodes in a full binary tree of height h is 2^h - 1.

A full binary tree of height 3 has 2^3 - 1 = 7 nodes. Inorder traversal visits all nodes exactly once.

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