Right Side View of Binary Tree in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare the struct for a binary tree node.

DSA C

typedef struct TreeNode {
    int val;
    struct TreeNode *[1];
    struct TreeNode *right;
} TreeNode;

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

Aleft

Bnode

Cchild

Droot

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'root' or 'node' instead of 'left' for the left child pointer.

The left child pointer in a binary tree node is usually named left.

2fill in blank

medium

Complete the code to initialize the queue for level order traversal.

DSA C

TreeNode* queue[100];
int front = 0, rear = 0;
queue[rear++] = [1];

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

Atemp

BNULL

Croot

Dnode

Attempts:

3 left

💡 Hint

Common Mistakes

Adding NULL or an uninitialized variable to the queue.

We start the level order traversal by adding the root node to the queue.

3fill in blank

hard

Fix the error in the condition to check if the queue is empty.

DSA C

while (front [1] rear) {
    TreeNode* node = queue[front++];
    // process node
}

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

A>

B<

C==

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>' or '==' which causes the loop to never run or run incorrectly.

The queue is not empty while front is less than rear.

4fill in blank

hard

Fill both blanks to add the right child to the queue if it exists.

DSA C

if (node->[1] != NULL) {
    queue[rear++] = node->[2];
}

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Aright

Bleft

Attempts:

3 left

💡 Hint

Common Mistakes

Checking or adding the left child instead of the right child.

We check if the right child exists and then add it to the queue.

5fill in blank

hard

Fill all three blanks to store the rightmost node's value at each level.

DSA C

int rightSideView[100];
int size = 0;
while (front < rear) {
    int levelSize = rear - front;
    for (int i = 0; i < levelSize; i++) {
        TreeNode* node = queue[front++];
        if (i == [1] - 1) {
            rightSideView[size++] = node->[2];
        }
        if (node->[3] != NULL) queue[rear++] = node->right;
        if (node->left != NULL) queue[rear++] = node->left;
    }
}

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

AlevelSize

Bval

Cright

Dleft

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong index for rightmost node or wrong child pointers.

The rightmost node is at index levelSize - 1, we store its value, and add its right child to the queue.

Practice

1. What does the "Right Side View" of a binary tree represent?

easy

A. All leaf nodes of the tree

B. Nodes visible when the tree is viewed from the left side

C. Nodes visible when the tree is viewed from the right side

D. Nodes at the deepest level only

Solution

1. Step 1: Understand the view perspective

   The right side view shows nodes visible when looking at the tree from the right side.

2. Step 2: Identify visible nodes

   Only the rightmost node at each level is visible from the right side.

3. Final Answer:

   Nodes visible when the tree is viewed from the right side -> Option C

4. Quick Check:

   Right side view = rightmost nodes [OK]

Hint: Right side view shows rightmost nodes at each level [OK]

Common Mistakes:

• Confusing right side view with left side view
• Thinking it shows all leaf nodes
• Assuming it shows only deepest nodes

2. Which of the following C code snippets correctly declares a binary tree node structure for use in right side view algorithms?

easy

A. typedef struct Node { int val; struct Node* left; struct Node* right; } Node;

B. struct Node { int val; Node* left; Node* right; }; typedef Node;

C. typedef struct { int val; struct Node* left; struct Node* right; } Node;

D. struct Node { int val; struct Node left; struct Node right; }; typedef struct Node Node;

Solution

1. Step 1: Check struct declaration syntax

   typedef struct Node { int val; struct Node* left; struct Node* right; } Node; correctly declares a struct with pointers to left and right child nodes.

2. Step 2: Verify typedef usage

   typedef struct Node { int val; struct Node* left; struct Node* right; } Node; uses typedef properly to alias struct Node as Node.

3. Final Answer:

   typedef struct Node { int val; struct Node* left; struct Node* right; } Node; -> Option A

4. Quick Check:

   Correct struct with pointers and typedef [OK]

Hint: Use struct with pointers and typedef for node [OK]

Common Mistakes:

• Missing struct keyword before pointer types
• Using struct members as values instead of pointers
• Incorrect typedef syntax

3. Given the following binary tree and code snippet, what is the output of the right side view function?

    1
   / \
  2   3
   \   \
    5   4

// Function call: rightSideView(root);

medium

A. [1, 2, 4]

B. [1, 2, 5]

C. [1, 3, 5]

D. [1, 3, 4]

Solution

1. Step 1: Identify rightmost nodes at each level

   Level 0: Node 1; Level 1: Node 3; Level 2: Node 4.

2. Step 2: Collect nodes in order

   The right side view list is [1, 3, 4].

3. Final Answer:

   [1, 3, 4] -> Option D

4. Quick Check:

   Rightmost nodes per level = [1, 3, 4] [OK]

Hint: Pick rightmost node at each tree level [OK]

Common Mistakes:

• Choosing left child nodes instead of rightmost
• Including nodes not visible from right side
• Confusing node values at levels

4. The following C function attempts to print the right side view of a binary tree but has a bug. What is the error?

void rightSideViewHelper(Node* root, int level, int* maxLevel) {
    if (!root) return;
    if (level > *maxLevel) {
        printf("%d ", root->val);
        *maxLevel = level;
    }
    rightSideViewHelper(root->left, level + 1, maxLevel);
    rightSideViewHelper(root->right, level + 1, maxLevel);
}

medium

A. The function visits left child before right, so right side nodes may be missed

B. The maxLevel pointer is not updated correctly

C. The base case does not check for null root

D. The printf statement is missing a newline character

Solution

1. Step 1: Analyze traversal order

   The function visits left child before right child, so rightmost nodes may be overshadowed by left nodes.

2. Step 2: Understand impact on right side view

   Right side view requires visiting right child first to capture rightmost nodes at each level.

3. Final Answer:

   The function visits left child before right, so right side nodes may be missed -> Option A

4. Quick Check:

   Visit right child first for correct right side view [OK]

Hint: Traverse right child before left child [OK]

Common Mistakes:

• Visiting left child before right child
• Not updating maxLevel properly
• Ignoring base case for null nodes

5. You want to modify a function that returns the right side view of a binary tree to also return the left side view in a single traversal. Which approach is best?

hard

A. Perform two separate traversals: one right-first for right view, one left-first for left view

B. Use a single traversal visiting both children and track first node at each level for left and last for right

C. Use a single traversal visiting right child first, store rightmost nodes, then visit left child for leftmost nodes

D. Traverse only left child and assume it covers both views

Solution

1. Step 1: Understand views require different nodes

   Right view needs rightmost nodes; left view needs leftmost nodes at each level.

2. Step 2: Use single traversal with tracking

   Track first node seen at each level for left view and last node for right view during one traversal.

3. Step 3: Avoid separate traversals for efficiency

   This approach reduces time by combining both views in one pass.

4. Final Answer:

   Use a single traversal visiting both children and track first node at each level for left and last for right -> Option B

5. Quick Check:

   Track first and last nodes per level in one traversal [OK]

Hint: Track first and last nodes per level in one traversal [OK]

Common Mistakes:

• Doing two full traversals unnecessarily
• Visiting only one child and missing nodes
• Not tracking nodes per level properly