Right Side View of Binary Tree in DSA C - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Right Side View for a Simple Binary Tree

What is the output of the right side view function for the given binary tree?

DSA C

struct TreeNode {
    int val;
    struct TreeNode *left;
    struct TreeNode *right;
};

// Tree structure:
//      1
//     / \
//    2   3
//     \   \
//      5   4

// Right side view output expected: [1, 3, 4]

A [1, 3, 5]

B [1, 3, 4]

C [1, 2, 4]

D [1, 2, 5]

💡 Hint

The right side view shows the rightmost node at each level.

At level 1, node 1 is visible. At level 2, node 3 is visible (right child). At level 3, node 4 is visible (right child of 3).

🧠 Conceptual

intermediate

Understanding Right Side View Node Selection

In the right side view of a binary tree, which nodes are included in the output?

A All leaf nodes

B The leftmost node at each level

C The rightmost node at each level

D All nodes with only one child

💡 Hint

Think about what you see when looking at the tree from the right side.

The right side view includes the rightmost node visible at each level of the tree.

❓ Predict Output

advanced

Output of Right Side View for Unbalanced Tree

What is the output of the right side view function for this unbalanced binary tree?

DSA C

// Tree structure:
//       10
//      /  
//     5    
//      \    
//       7   
//        \
//         9

// Expected output: [10, 5, 7, 9]

A [10, 5, 7, 9]

B [10, 5, 9]

C [10, 7, 9]

D [10, 5, 7]

💡 Hint

Right side view includes the rightmost node at each depth level.

At level 1: 10; level 2: 5 (only node); level 3: 7 (right child of 5); level 4: 9 (right child of 7).

🔧 Debug

advanced

Identify the Error in Right Side View Implementation

What error will this code produce when trying to get the right side view of a binary tree?

DSA C

void rightSideViewHelper(struct TreeNode* root, int level, int* maxLevel, int* result, int* returnSize) {
    if (!root) return;
    if (level > *maxLevel) {
        result[(*returnSize)++] = root->val;
        *maxLevel = level;
    }
    rightSideViewHelper(root->right, level + 1, maxLevel, result, returnSize);
    rightSideViewHelper(root->left, level + 1, maxLevel, result, returnSize);
}

// The helper visits right child before left child.

A The code runs correctly and outputs the right side view

B The code causes a segmentation fault

C The output includes duplicate nodes

D The output misses rightmost nodes at each level

💡 Hint

Check the order of traversal in the helper function.

Visiting right child before left child causes the first node at each level to be the rightmost, so right side nodes are captured correctly.

🚀 Application

expert

Number of Nodes in Right Side View for Complete Binary Tree

Given a complete binary tree with height h (root at height 1), how many nodes will the right side view contain?

A 2^h - 1

B 2^(h-1)

C h + 1

D h

💡 Hint

Right side view shows one node per level.

In a complete binary tree, the right side view shows exactly one node per level, so the count equals the height h.

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