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

Practice - 5 Tasks

1fill in blank

easy

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

DSA C

void inorderTraversal(struct Node* root) {
    if (root == NULL) return;
    inorderTraversal(root->[1]);
    printf("%d ", root->data);
    inorderTraversal(root->right);
}

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

Achild

Bright

Cparent

Dleft

Attempts:

3 left

💡 Hint

Common Mistakes

Using root->right instead of root->left.

Trying to access root->parent which does not exist in this structure.

In inorder traversal, we first visit the left child, so root->left is correct.

2fill in blank

medium

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

DSA C

void inorderTraversal(struct Node* root) {
    if (root == NULL) return;
    inorderTraversal(root->left);
    printf("%[1] ", root->data);
    inorderTraversal(root->right);
}

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

Af

Bd

Cs

Dc

Attempts:

3 left

💡 Hint

Common Mistakes

Using %f which is for floats.

Using %s which is for strings.

The format specifier for integers in printf is %d.

3fill in blank

hard

Fix the error in the recursive call to traverse the right subtree in inorder traversal.

DSA C

void inorderTraversal(struct Node* root) {
    if (root == NULL) return;
    inorderTraversal(root->left);
    printf("%d ", root->data);
    inorderTraversal(root->[1]);
}

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

Aright

Bchild

Cparent

Dleft

Attempts:

3 left

💡 Hint

Common Mistakes

Using root->left again instead of root->right.

Trying to access root->parent which is not part of this traversal.

After visiting the root, inorder traversal visits the right subtree using root->right.

4fill in blank

hard

Fill both blanks to complete the inorder traversal function correctly.

DSA C

void inorderTraversal(struct Node* root) {
    if (root == NULL) return;
    inorderTraversal(root->[1]);
    printf("%d ", root->data);
    inorderTraversal(root->[2]);
}

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

Aleft

Bright

Cparent

Dchild

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping left and right children.

Using non-existent pointers like parent or child.

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

5fill in blank

hard

Fill all three blanks to create a function that returns the inorder traversal as a string of node values separated by spaces.

DSA C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

void inorderTraversalStr(struct Node* root, char* result) {
    if (root == NULL) return;
    inorderTraversalStr(root->[1], result);
    char buffer[12];
    sprintf(buffer, "%[2] ", root->data);
    strcat(result, buffer);
    inorderTraversalStr(root->[3], result);
}

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

Aleft

Bd

Cright

Df

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong pointers like parent or child.

Using incorrect format specifiers like %f or %s.

The function visits left child, appends the root data with %d format, then visits right child.

Practice

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

easy

A. Left subtree, Root, Right subtree

B. Root, Left subtree, Right subtree

C. Right subtree, Root, Left subtree

D. Root, Right subtree, Left subtree

Solution

1. Step 1: Understand inorder traversal definition

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

2. Step 2: Apply the order to the tree

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

3. Final Answer:

   Left subtree, Root, Right subtree -> Option A

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 C function prototypes correctly represents an inorder traversal of a binary tree node struct Node?

easy

A. int inorder(struct Node* root);

B. int inorder(struct Node root);

C. void inorder(struct Node root);

D. void inorder(struct Node* root);

Solution

1. Step 1: Identify correct parameter type for tree traversal

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

2. Step 2: Check return type and parameter

   Traversal functions typically return void as they print or process nodes during traversal, so void return type with pointer parameter is correct.

3. Final Answer:

   void inorder(struct Node* root); -> Option D

4. Quick Check:

   Traversal uses pointer and void return [OK]

Hint: Traversal functions use pointer to root and void return [OK]

Common Mistakes:

• Passing struct by value instead of pointer
• Using int return type instead of void
• Missing pointer in parameter

3. Given the binary tree below:

    2
   / \
  1   3

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

medium

A. 2 1 3

B. 1 3 2

C. 1 2 3

D. 3 2 1

Solution

1. Step 1: Identify inorder traversal order

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

2. Step 2: Apply inorder to given tree

   Left child is 1, root is 2, right child is 3, so output is "1 2 3".

3. Final Answer:

   1 2 3 -> Option C

4. Quick Check:

   Inorder of 2-node tree = 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 code snippet for inorder traversal:

void inorder(struct Node* root) {
    if (root == NULL) return;
    inorder(root->left);
    printf("%d ", root->data);
    inorder(root->right);
}

What is the error if the function prints nodes in preorder instead of inorder?

medium

A. The printf statement is before the left subtree call

B. The base case is missing the NULL check

C. The right subtree call is missing

D. The function is missing a return type

Solution

1. Step 1: Understand traversal order in code

   Inorder prints after left subtree call and before right subtree call.

2. Step 2: Identify cause of preorder output

   If printf is before left subtree call, root prints first, causing preorder output.

3. Final Answer:

   The printf statement is before the left subtree call -> Option A

4. Quick Check:

   Print before left subtree = preorder [OK]

Hint: Print root between left and right calls for inorder [OK]

Common Mistakes:

• Placing print before left subtree call
• Omitting NULL base case
• Forgetting to call right subtree

5. You have a binary search tree with nodes inserted in this order: 5, 3, 7, 2, 4, 6, 8.
What is the inorder traversal output of this tree?

hard

A. 5 3 7 2 4 6 8

B. 2 3 4 5 6 7 8

C. 8 7 6 5 4 3 2

D. 3 2 4 5 6 7 8

Solution

1. Step 1: Build the BST structure

   Insert 5 as root, 3 left of 5, 7 right of 5, 2 left of 3, 4 right of 3, 6 left of 7, 8 right of 7.

2. Step 2: Perform inorder traversal

   Inorder visits nodes in ascending order for BST: 2, 3, 4, 5, 6, 7, 8.

3. Final Answer:

   2 3 4 5 6 7 8 -> Option B

4. Quick Check:

   Inorder of BST = sorted nodes [OK]

Hint: Inorder traversal of BST gives sorted order [OK]

Common Mistakes:

• Listing nodes in insertion order
• Reversing the order
• Skipping nodes in traversal