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

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare the vector that will store the right side view nodes.

DSA C++

std::vector<int> [1];

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Aresult

BrightView

Coutput

DsideNodes

Attempts:

3 left

💡 Hint

Common Mistakes

Using a variable name that is not declared or inconsistent with the rest of the code.

The vector named 'result' is commonly used to store the right side view nodes in this implementation.

2fill in blank

medium

Complete the code to check if the current node is null before processing.

DSA C++

if ([1] == nullptr) return;

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Anode

Bcurrent

Croot

Dtemp

Attempts:

3 left

💡 Hint

Common Mistakes

Checking a variable that is not the current node pointer.

The parameter 'node' is used to represent the current node in the recursive function, so we check if 'node' is null.

3fill in blank

hard

Fix the error in the condition that adds the node's value to the result vector only if it's the first node at this level.

DSA C++

if (level == [1]) result.push_back(node->val);

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A0

Blevel

Cnode->val

Dresult.size()

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing level with level itself or with node value instead of result size.

We compare the current level with the size of the result vector to ensure only the first node at each level is added.

4fill in blank

hard

Fill both blanks to recursively traverse the right child first, then the left child.

DSA C++

dfs([1], level + 1, result);
dfs([2], level + 1, result);

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Anode->right

Bnode->left

Croot->right

Droot->left

Attempts:

3 left

💡 Hint

Common Mistakes

Visiting left child before right child, which gives wrong view.

To get the right side view, we visit the right child first (node->right), then the left child (node->left).

5fill in blank

hard

Fill all three blanks to complete the main function that calls the helper and returns the right side view.

DSA C++

std::vector<int> rightSideView(TreeNode* root) {
    std::vector<int> [1];
    dfs([2], 0, [3]);
    return [3];
}

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

Aresult

Broot

Doutput

Attempts:

3 left

💡 Hint

Common Mistakes

Using inconsistent variable names or forgetting to return the result vector.

We declare 'result' vector, call dfs with root and level 0, then return 'result'.

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