Right Side View of Binary Tree in DSA C++ - Execution Trace

Concept Flow - Right Side View of Binary Tree

Start at root node

↓

Use queue for level order traversal

↓

For each level:

↓

Traverse all nodes

→Record last node's value

↓

Add children to queue

↓

Repeat until queue empty

↓

Return collected right side view

We visit the tree level by level, always noting the last node at each level to get the right side view.

Execution Sample

DSA C++

vector<int> rightSideView(TreeNode* root) {
  vector<int> result;
  if (!root) return result;
  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;
}

This code collects the rightmost node's value at each level of the binary tree using a queue.

Execution Table

Step	Operation	Nodes in Queue	Node Processed	Pointer Changes	Right Side View Collected	Visual Tree State
1	Initialize queue with root	[1]	None	q = [1]	[]	1 / \ 2 3 / \ 4 5
2	Process level 1	[1]	1	Pop 1, push 2 and 3	[1]	1 / \ 2 3 / \ 4 5
3	Process level 2	[2,3]	2	Pop 2, push 4	[1,3]	1 / \ 2 3 / \ 4 5
4	Process level 2	[3,4]	3	Pop 3, push 5	[1,3,5]	1 / \ 2 3 / \ 4 5
5	Process level 3	[4,5]	4	Pop 4, no children	[1,3,5]	1 / \ 2 3 / \ 4 5
6	Process level 3	[5]	5	Pop 5, no children	[1,3,5]	1 / \ 2 3 / \ 4 5
7	Queue empty, end	[]	None	None	[1,3,5]	1 / \ 2 3 / \ 4 5

💡 Queue is empty, all levels processed, right side view collected.

Variable Tracker

Variable	Start	After Step 2	After Step 4	After Step 6	Final
q (queue)	[]	[2,3]	[4,5]	[]	[]
result (right side view)	[]	[1]	[1,3]	[1,3,5]	[1,3,5]
node (current processed)	None	1	3	5	None

Key Moments - 3 Insights

Why do we record the node value only when i == size - 1 in the loop?

Because i == size - 1 means the node is the last in the current level, which is the rightmost node visible from the right side. See execution_table rows 3 and 4 where nodes 1 and 3 are recorded.

Why do we use a queue for this traversal?

A queue helps us process nodes level by level (breadth-first). This ensures we visit all nodes at one level before moving to the next, which is essential to identify the rightmost node at each level. See execution_table steps 2 to 6 showing queue changes.

What happens if the tree is empty?

If the root is null, the queue stays empty and the function returns an empty result immediately. This is shown at step 1 where the queue is initialized and checked.

Visual Quiz - 3 Questions

Look at the execution_table at Step 4, which node is processed and what is added to the queue?

A Node 2 processed, children 4 added

B Node 3 processed, no children added

C Node 3 processed, child 5 added

D Node 1 processed, children 2 and 3 added

💡 Hint

Check 'Node Processed' and 'Pointer Changes' columns at Step 4 in execution_table.

At which step is the value 5 added to the right side view?

A Step 5

B Step 6

C Step 4

D Step 7

💡 Hint

Look at the 'Right Side View Collected' column in execution_table rows 5 and 6.

If the tree had no right child at the root, how would the queue look after Step 2?

A [2]

B [3]

C []

D [2,3]

💡 Hint

Refer to the 'Pointer Changes' column at Step 2 and imagine no right child to add.

Concept Snapshot

Right Side View of Binary Tree:
- Use level order traversal with a queue
- For each level, record the last node's value
- Add left and right children to queue
- Continue until queue is empty
- Result is list of rightmost nodes per level

Full Transcript

To find the right side view of a binary tree, we visit nodes level by level using a queue. At each level, we process all nodes and record the value of the last node processed. We add the children of each node to the queue for the next level. This way, the collected values represent the nodes visible from the right side. If the tree is empty, we return an empty list immediately. The process ends when the queue is empty, meaning all levels have been processed.

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