Right Side View of Binary Tree in DSA C++ - Build from Scratch

Right Side View of Binary Tree

📖 Scenario: Imagine you have a tree of family members, and you want to see only the members visible from the right side. This means you want to see the rightmost member at each level of the tree.

🎯 Goal: Build a program that shows the right side view of a binary tree by printing the values of the nodes visible from the right side.

📋 What You'll Learn

Create a binary tree with the exact structure given

Use a variable to track the current level during traversal

Implement a function to find the right side view of the tree

Print the right side view nodes in order

💡 Why This Matters

🌍 Real World

Right side view of a binary tree helps in visualizing hierarchical data from a specific perspective, useful in UI trees, organizational charts, and decision trees.

💼 Career

Understanding tree traversal and views is important for software engineers working with data structures, algorithms, and system design.

1

Create the Binary Tree Structure

Create a binary tree with the exact structure below using the TreeNode struct. The root node has value 1, its left child is 2, right child is 3. Node 2 has a right child 5, and node 3 has a right child 4. Use pointers named exactly as root, node2, node3, node5, and node4.

DSA C++

#include 
using namespace std;

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

int main() {
    // Create the binary tree nodes and link them
    // Your code here
    return 0;
}

Hint

Start by creating nodes with new TreeNode(value). Then link them using ->left and ->right.

2

Add a Variable to Track Current Level

Add an integer variable called maxLevel and set it to 0. This will help track the deepest level visited during traversal.

DSA C++

#include 
using namespace std;

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

int main() {
    TreeNode* root = new TreeNode(1);
    TreeNode* node2 = new TreeNode(2);
    TreeNode* node3 = new TreeNode(3);
    TreeNode* node5 = new TreeNode(5);
    TreeNode* node4 = new TreeNode(4);

    root->left = node2;
    root->right = node3;
    node2->right = node5;
    node3->right = node4;

    // Add maxLevel variable here
    // Your code here
    return 0;
}

Hint

Declare int maxLevel = 0; before traversal starts.

3

Implement the Right Side View Function

Write a recursive function called rightView that takes TreeNode* node, int level, int& maxLevel, and prints the node's value if level is greater than maxLevel. Update maxLevel accordingly. Traverse the right child first, then the left child.

DSA C++

#include 
using namespace std;

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

void rightView(TreeNode* node, int level, int& maxLevel) {
    // Implement the recursive rightView function here
    // Your code here
}

int main() {
    TreeNode* root = new TreeNode(1);
    TreeNode* node2 = new TreeNode(2);
    TreeNode* node3 = new TreeNode(3);
    TreeNode* node5 = new TreeNode(5);
    TreeNode* node4 = new TreeNode(4);

    root->left = node2;
    root->right = node3;
    node2->right = node5;
    node3->right = node4;

    int maxLevel = 0;

    return 0;
}

Hint

Check if node is null first. Then if level > maxLevel, print the node value and update maxLevel. Recurse right child first, then left child.

4

Print the Right Side View

Call the rightView function from main with root, starting level as 1, and the maxLevel variable. Then print a newline.

DSA C++

#include 
using namespace std;

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

void rightView(TreeNode* node, int level, int& maxLevel) {
    if (node == nullptr) {
        return;
    }
    if (level > maxLevel) {
        cout val right, level + 1, maxLevel);
    rightView(node->left, level + 1, maxLevel);
}

int main() {
    TreeNode* root = new TreeNode(1);
    TreeNode* node2 = new TreeNode(2);
    TreeNode* node3 = new TreeNode(3);
    TreeNode* node5 = new TreeNode(5);
    TreeNode* node4 = new TreeNode(4);

    root->left = node2;
    root->right = node3;
    node2->right = node5;
    node3->right = node4;

    int maxLevel = 0;

    // Call rightView function and print newline
    // Your code here
    return 0;
}

Hint

Call rightView(root, 1, maxLevel); and then print a newline with cout << endl;.

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