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;
    TreeNode* left;
    TreeNode* right;
    TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
};

#include <vector>
#include <queue>
#include <iostream>

std::vector<int> rightSideView(TreeNode* root) {
    std::vector<int> result;
    if (!root) return result;
    std::queue<TreeNode*> q;
    q.push(root);
    while (!q.empty()) {
        int size = q.size();
        for (int i = 0; i < size; ++i) {
            TreeNode* node = q.front();
            q.pop();
            if (i == size - 1) result.push_back(node->val);
            if (node->left) q.push(node->left);
            if (node->right) q.push(node->right);
        }
    }
    return result;
}

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

int main() {
    TreeNode* root = new TreeNode(1);
    root->left = new TreeNode(2);
    root->right = new TreeNode(3);
    root->left->right = new TreeNode(5);
    root->right->right = new TreeNode(4);

    std::vector<int> output = rightSideView(root);
    for (int val : output) {
        std::cout << val << " ";
    }
    return 0;
}

A [1, 3, 4]

B [1, 2, 5]

C [1, 3, 5]

D [1, 2, 4]

💡 Hint

Think about which nodes are visible when looking from the right side level by level.

The right side view shows the last node at each level when traversed from left to right. At level 1, node 1 is visible; at level 2, node 3; at level 3, node 4.

🧠 Conceptual

intermediate

Understanding the Right Side View Algorithm

Which of the following best describes the approach used to find the right side view of a binary tree?

A Traverse the tree level by level and record the last node at each level.

B Traverse the tree level by level and record the first node at each level.

C Traverse the tree in preorder and record all right children.

D Traverse the tree in inorder and record nodes with no left child.

💡 Hint

Consider which node is visible from the right side at each level.

The right side view is obtained by recording the last node visited at each level during a level order traversal.

🔧 Debug

advanced

Identify the Error in Right Side View Implementation

What error will occur when running the following code snippet for right side view?

DSA C++

std::vector<int> rightSideView(TreeNode* root) {
    std::vector<int> result;
    std::queue<TreeNode*> q;
    q.push(root);
    while (!q.empty()) {
        int size = q.size();
        for (int i = 0; i <= size; ++i) {
            TreeNode* node = q.front();
            q.pop();
            if (i == size - 1) result.push_back(node->val);
            if (node->left) q.push(node->left);
            if (node->right) q.push(node->right);
        }
    }
    return result;
}

A Logic error: returns empty vector

B Runtime error due to popping from empty queue

C Compilation error due to missing semicolon

D No error, returns correct output

💡 Hint

Check the loop boundary condition carefully.

The for loop uses i <= size which causes one extra iteration, leading to popping from an empty queue and runtime error.

❓ Predict Output

advanced

Output of Right Side View for Left-Skewed Tree

What is the output of the right side view function for this left-skewed binary tree?

DSA C++

// Tree structure:
//     1
//    /
//   2
//  /
// 3
//
#include <iostream>
#include <vector>
#include <queue>

struct TreeNode {
    int val;
    TreeNode* left;
    TreeNode* right;
    TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
};

std::vector<int> rightSideView(TreeNode* root) {
    std::vector<int> result;
    if (!root) return result;
    std::queue<TreeNode*> q;
    q.push(root);
    while (!q.empty()) {
        int size = q.size();
        for (int i = 0; i < size; ++i) {
            TreeNode* node = q.front();
            q.pop();
            if (i == size - 1) result.push_back(node->val);
            if (node->left) q.push(node->left);
            if (node->right) q.push(node->right);
        }
    }
    return result;
}

int main() {
    TreeNode* root = new TreeNode(1);
    root->left = new TreeNode(2);
    root->left->left = new TreeNode(3);

    std::vector<int> output = rightSideView(root);
    for (int val : output) {
        std::cout << val << " ";
    }
    return 0;
}

A [3]

B [1, 3]

C [1]

D [1, 2, 3]

💡 Hint

Even if the tree is left-skewed, the right side view shows the rightmost node at each level.

Since there are no right children, the right side view shows the only node at each level: 1, then 2, then 3.

🚀 Application

expert

Number of Nodes in Right Side View for Complete Binary Tree

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

A h - 1

B 2^h - 1

C h

D 2^(h-1)

💡 Hint

Right side view shows one node per level.

The right side view contains exactly one node per level, so for height h, it contains h nodes.

Practice

1. What does the "Right Side View of Binary Tree" represent?

easy

A. Nodes at the deepest level of the tree

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

C. All leaf nodes of the binary tree

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

Solution

1. Step 1: Understand the view perspective

   The right side view shows nodes visible when looking at the tree from the right side, ignoring nodes hidden behind others.

2. Step 2: Differentiate from other views

   Unlike left side or leaf views, right side view specifically captures the rightmost nodes at each level.

3. Final Answer:

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

4. Quick Check:

   Right side view = rightmost nodes [OK]

Hint: Right side view shows last node at each tree level [OK]

Common Mistakes:

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

2. Which of the following C++ code snippets correctly declares a function to get the right side view of a binary tree?

easy

A. vector<int> rightSideView(TreeNode* root);

B. void rightSideView(TreeNode root);

C. int rightSideView(TreeNode* root);

D. vector<int> rightSideView(TreeNode root);

Solution

1. Step 1: Check function return type

   The right side view returns a list of node values, so vector<int> is appropriate.

2. Step 2: Check parameter type

   Function should take a pointer to root node (TreeNode*), not a value copy.

3. Final Answer:

   vector<int> rightSideView(TreeNode* root); -> Option A

4. Quick Check:

   Return vector<int>, parameter TreeNode* [OK]

Hint: Use vector<int> return and TreeNode* parameter [OK]

Common Mistakes:

• Passing TreeNode by value instead of pointer
• Returning int instead of vector<int>
• Using void return type incorrectly

3. Given the binary tree below, what is the right side view output?

    1
   / \
  2   3
   \   \
    5   4

medium

A. [1, 3, 4]

B. [1, 2, 5]

C. [1, 3, 5]

D. [1, 2, 4]

Solution

1. Step 1: Identify rightmost nodes at each level

   Level 1: Node 1; Level 2: Node 3 (right child of 1); Level 3: Node 4 (right child of 3).

2. Step 2: Collect nodes in order

   Right side view is [1, 3, 4].

3. Final Answer:

   [1, 3, 4] -> Option A

4. Quick Check:

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

Hint: Pick last node visible at each tree level from right [OK]

Common Mistakes:

• Choosing left child nodes instead of rightmost
• Including nodes hidden behind others
• Mixing node order in output

4. Consider this C++ function snippet to get the right side view:

void dfs(TreeNode* node, int level, vector<int>& res) {
  if (!node) return;
  if (level == res.size()) res.push_back(node->val);
  dfs(node->left, level + 1, res);
  dfs(node->right, level + 1, res);
}

What is the main bug in this code?

medium

A. It does not check for null nodes

B. It uses pass-by-value for vector, causing no updates

C. It pushes duplicate values into result vector

D. It visits left child before right, so rightmost nodes may be missed

Solution

1. Step 1: Analyze traversal order

   Function visits left child before right child, so first node at each level is leftmost, not rightmost.

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:

   It visits left child before right, so rightmost nodes may be missed -> Option D

4. Quick Check:

   Right side view needs right-first DFS [OK]

Hint: Visit right child before left for correct right side view [OK]

Common Mistakes:

• Visiting left child first in DFS for right side view
• Ignoring null node checks (though present here)
• Passing vector by value instead of reference

5. You want to find the right side view of a binary tree but the tree is very deep and unbalanced. Which approach is best to avoid stack overflow and get the correct right side view?

hard

A. Use recursive DFS visiting right child first without depth limit

B. Use recursive DFS visiting left child first

C. Use iterative level order traversal with a queue, capturing last node at each level

D. Use inorder traversal and collect nodes

Solution

1. Step 1: Consider recursion limits on deep trees

   Deep unbalanced trees risk stack overflow with recursion, so iterative methods are safer.

2. Step 2: Choose traversal method for right side view

   Level order traversal with a queue visits nodes level by level; capturing last node per level gives right side view.

3. Step 3: Evaluate other options

   Recursive DFS may cause stack overflow; inorder traversal does not capture right side view correctly.

4. Final Answer:

   Use iterative level order traversal with a queue, capturing last node at each level -> Option C

5. Quick Check:

   Iterative BFS with queue = safe & correct [OK]

Hint: Use BFS queue, pick last node per level to avoid recursion issues [OK]

Common Mistakes:

• Using recursion on very deep trees causing stack overflow
• Visiting left child first in DFS for right side view
• Using inorder traversal which doesn't capture right side view