Tree Traversal Inorder Left Root Right in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to print the left subtree first in inorder traversal.

DSA C++

void inorderTraversal(Node* root) {
    if (root == nullptr) return;
    inorderTraversal(root->[1]);
    std::cout << root->data << " ";
    inorderTraversal(root->right);
}

Drag options to blanks, or click blank then click option'

Aleft

Bright

Cparent

Dchild

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'right' instead of 'left' causes the traversal to visit the right subtree first.

Using 'parent' or 'child' is incorrect because those are not valid pointers in this context.

In inorder traversal, we first visit the left child, then the root, and finally the right child.

2fill in blank

medium

Complete the code to print the root node's data in inorder traversal.

DSA C++

void inorderTraversal(Node* root) {
    if (root == nullptr) return;
    inorderTraversal(root->left);
    std::cout << root->[1] << " ";
    inorderTraversal(root->right);
}

Drag options to blanks, or click blank then click option'

Aleft

Bdata

Cright

Dvalue

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'left' or 'right' instead of 'data' prints pointers, not values.

Using 'value' is incorrect if the struct uses 'data' as the member name.

The root's data is printed between visiting the left and right subtrees in inorder traversal.

3fill in blank

hard

Fix the error in the inorder traversal function to correctly visit the right subtree.

DSA C++

void inorderTraversal(Node* root) {
    if (root == nullptr) return;
    inorderTraversal(root->left);
    std::cout << root->data << " ";
    inorderTraversal(root->[1]);
}

Drag options to blanks, or click blank then click option'

Aleft

Bchild

Cparent

Dright

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'left' again causes infinite recursion or incorrect traversal.

Using 'parent' or 'child' is invalid in this context.

After visiting the root, inorder traversal visits the right subtree, so the pointer should be 'right'.

4fill in blank

hard

Fill both blanks to complete the inorder traversal function that prints nodes in left-root-right order.

DSA C++

void inorderTraversal(Node* root) {
    if (root == nullptr) return;
    inorderTraversal(root->[1]);
    std::cout << root->[2] << " ";
    inorderTraversal(root->right);
}

Drag options to blanks, or click blank then click option'

Aleft

Bdata

Cright

Dvalue

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping left and right pointers.

Using 'value' instead of 'data' if the struct uses 'data'.

In inorder traversal, visit left child first, then print the root's data.

5fill in blank

hard

Fill all three blanks to complete the inorder traversal function visiting left, root, then right.

DSA C++

void inorderTraversal(Node* root) {
    if (root == nullptr) return;
    inorderTraversal(root->[1]);
    std::cout << root->[2] << " ";
    inorderTraversal(root->[3]);
}

Drag options to blanks, or click blank then click option'

Aleft

Bdata

Cright

Dvalue

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'value' instead of 'data' for the node's value.

Mixing up left and right pointers.

The inorder traversal visits the left child, then prints the root's data, then visits the right child.

Practice

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

easy

A. Right subtree, Root, Left subtree

B. Root, Left subtree, Right subtree

C. Left subtree, Root, Right subtree

D. Root, Right subtree, Left subtree

Solution

1. Step 1: Understand inorder traversal definition

   Inorder traversal visits nodes in the order: left child, then root, then right child.

2. Step 2: Apply to binary tree traversal

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

3. Final Answer:

   Left subtree, Root, Right subtree -> Option C

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 is the correct C++ function signature for inorder traversal of a binary tree node pointer Node* root?

easy

A. int inorderTraversal(Node* root)

B. int inorderTraversal(Node root)

C. void inorderTraversal(Node root)

D. void inorderTraversal(Node* root)

Solution

1. Step 1: Identify correct parameter type

   Traversal functions usually take a pointer to the root node to access the tree nodes.

2. Step 2: Identify correct return type

   Traversal functions typically return void because they print or process nodes without returning a value.

3. Final Answer:

   void inorderTraversal(Node* root) -> Option D

4. Quick Check:

   Pointer parameter and void return [OK]

Hint: Traversal functions use Node* and void return [OK]

Common Mistakes:

• Passing node by value instead of pointer
• Using int return type instead of void
• Missing pointer symbol * in parameter

3. Given the binary tree:

    2
   / \
  1   3

What is the output of the inorder traversal function that prints node values separated by spaces?

medium

A. 1 2 3

B. 2 1 3

C. 3 2 1

D. 1 3 2

Solution

1. Step 1: Traverse left subtree

   Start at root (2), go left to node (1), print 1 since it has no left child.

2. Step 2: Visit root and right subtree

   Return to root (2), print 2, then go right to node (3), print 3.

3. Final Answer:

   1 2 3 -> Option A

4. Quick Check:

   Inorder traversal output = 1 2 3 [OK]

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

Common Mistakes:

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

4. Consider this C++ inorder traversal code snippet:

void inorder(Node* root) {
  if (root == nullptr) return;
  inorder(root->left);
  std::cout << root->val << " ";
  inorder(root->right);
}

What is the error if the output is empty for a non-empty tree?

medium

A. The base case condition is incorrect

B. The function is not called with the root node

C. The print statement is missing

D. The traversal order is wrong

Solution

1. Step 1: Check base case correctness

   The base case correctly returns when root is nullptr, so no error here.

2. Step 2: Verify function call

   If output is empty, likely the function was never called with the actual root node of the tree.

3. Final Answer:

   The function is not called with the root node -> Option B

4. Quick Check:

   Missing initial call causes no output [OK]

Hint: Always call traversal with actual root node [OK]

Common Mistakes:

• Not calling traversal function at all
• Calling with nullptr instead of root
• Assuming print statement missing when it exists

5. You have a binary tree with nodes:

      4
     / \
    2   6
   / \ / \
  1  3 5  7

Using inorder traversal, which sequence of node values will you get?

hard

A. [1, 2, 3, 4, 5, 6, 7]

B. [4, 2, 1, 3, 6, 5, 7]

C. [7, 6, 5, 4, 3, 2, 1]

D. [1, 3, 2, 5, 7, 6, 4]

Solution

1. Step 1: Traverse left subtree of root (4)

   Visit left subtree rooted at 2: inorder visits 1, then 2, then 3.

2. Step 2: Visit root and right subtree

   After left subtree, visit root 4, then right subtree rooted at 6: visit 5, then 6, then 7.

3. Final Answer:

   [1, 2, 3, 4, 5, 6, 7] -> Option A

4. Quick Check:

   Inorder traversal of BST gives sorted sequence [OK]

Hint: Inorder traversal of BST returns sorted node values [OK]

Common Mistakes:

• Mixing preorder or postorder sequences
• Visiting root before left subtree
• Visiting right subtree before root