Right Side View of Binary Tree in DSA C++ - Time & Space Complexity

Time Complexity: Right Side View of Binary Tree

O(n)

Understanding Time Complexity

We want to understand how the time needed to find the right side view of a binary tree changes as the tree grows.

How does the number of nodes affect the work done?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

void rightSideViewHelper(TreeNode* root, int level, vector& view) {
    if (!root) return;
    if (level == view.size())
        view.push_back(root->val);
    rightSideViewHelper(root->right, level + 1, view);
    rightSideViewHelper(root->left, level + 1, view);
}

vector rightSideView(TreeNode* root) {
    vector view;
    rightSideViewHelper(root, 0, view);
    return view;
}
    

This code visits nodes to collect the rightmost node at each level of the tree.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive calls visiting each node once.
• How many times: Each node in the tree is visited exactly one time.

How Execution Grows With Input

As the number of nodes increases, the function visits each node once, so the work grows directly with the number of nodes.

Input Size (n)	Approx. Operations
10	About 10 visits
100	About 100 visits
1000	About 1000 visits

Pattern observation: The work grows linearly as the tree size grows.

Final Time Complexity

Time Complexity: O(n)

This means the time to find the right side view grows directly with the number of nodes in the tree.

Common Mistake

[X] Wrong: "Since we only look at the right side, the time is less than visiting all nodes."

[OK] Correct: The code still visits every node to check if it is the rightmost at its level, so all nodes are processed.

Interview Connect

Understanding this time complexity helps you explain how tree traversal works and shows you can analyze recursive solutions clearly.

Self-Check

"What if we changed the traversal to visit the left child before the right child? How would the time complexity change?"

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