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.