Right Side View of Binary Tree in DSA Javascript - 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 to see the tree from the right side?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function rightSideView(root) {
  if (!root) return [];
  const queue = [root];
  const result = [];
  while (queue.length > 0) {
    const size = queue.length;
    for (let i = 0; i < size; i++) {
      const node = queue.shift();
      if (node.left) queue.push(node.left);
      if (node.right) queue.push(node.right);
      if (i === size - 1) result.push(node.val);
    }
  }
  return result;
}
    

This code visits each node level by level and records the last node's value at each level to get the right side view.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop runs once per level, and inside it, the for loop visits every node at that level.
• How many times: Every node in the tree is visited exactly once during the for loops combined.

How Execution Grows With Input

As the number of nodes (n) grows, the code visits each node once, so the total operations grow roughly in direct proportion to n.

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

Pattern observation: The work grows linearly with the number of nodes.

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 take one node per level, the time complexity is O(height) or O(log n)."

[OK] Correct: Even though we record one node per level, we still visit every node to know which one is last at each level, so the time depends on all nodes, not just the height.

Interview Connect

Understanding this helps you explain how level-by-level tree traversal works and why visiting all nodes is necessary, a common pattern in tree problems.

Self-Check

"What if we used a depth-first search instead of breadth-first search? How would the time complexity change?"

Practice

1. What does the Right Side View of a binary tree represent?

easy

A. The nodes at the deepest level of the tree

B. The nodes visible when the tree is viewed from the right side

C. All leaf nodes of the tree

D. The 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.

2. Step 2: Identify visible nodes

   Only the rightmost node at each level is visible from the right side.

3. Final Answer:

   The nodes visible when the tree is viewed from the right side -> Option B

4. Quick Check:

   Right side view = rightmost nodes at each level [OK]

Hint: Right side view shows rightmost nodes 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 JavaScript function signatures correctly represents a function to get the right side view of a binary tree?

easy

A. function rightSideView() { }

B. function rightSideView(root, result) { }

C. function rightSideView(root) { }

D. function rightSideView(tree, level) { }

Solution

1. Step 1: Check typical function signature

   The function usually takes the root node as input to start traversal.

2. Step 2: Validate options

   function rightSideView(root) { } correctly takes only the root node, which is standard for this problem.

3. Final Answer:

   function rightSideView(root) { } -> Option C

4. Quick Check:

   Function takes root node only [OK]

Hint: Function usually takes root node as single parameter [OK]

Common Mistakes:

• Adding unnecessary parameters
• Omitting root parameter
• Using unclear parameter names

3. What is the output of the following code snippet for the given binary tree?

class TreeNode {
  constructor(val, left = null, right = null) {
    this.val = val;
    this.left = left;
    this.right = right;
  }
}

const root = new TreeNode(1,
  new TreeNode(2, null, new TreeNode(5)),
  new TreeNode(3, null, new TreeNode(4))
);

function rightSideView(root) {
  const result = [];
  const queue = [root];
  while (queue.length) {
    const size = queue.length;
    for (let i = 0; i < size; i++) {
      const node = queue.shift();
      if (i === size - 1) result.push(node.val);
      if (node.left) queue.push(node.left);
      if (node.right) queue.push(node.right);
    }
  }
  return result;
}

console.log(rightSideView(root));

medium

A. [1, 3, 4]

B. [1, 2, 5]

C. [1, 3, 5]

D. [1, 2, 4]

Solution

1. Step 1: Traverse level by level and identify rightmost nodes

   Level 1: Node 1 (only node), Level 2: Nodes 2 and 3 (rightmost is 3), Level 3: Nodes 5 and 4 (rightmost is 4).

2. Step 2: Collect rightmost nodes in order

   Result array 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 at each level during BFS [OK]

Common Mistakes:

• Choosing leftmost nodes instead of rightmost
• Mixing node values from different levels
• Forgetting to check all nodes at a level

4. Identify the bug in the following code that attempts to find the right side view of a binary tree:

function rightSideView(root) {
  const result = [];
  const queue = [root];
  while (queue.length) {
    const size = queue.length;
    for (let i = 0; i < size; i++) {
      const node = queue.shift();
      if (i === 0) result.push(node.val);
      if (node.left) queue.push(node.left);
      if (node.right) queue.push(node.right);
    }
  }
  return result;
}

medium

A. It pushes the leftmost node instead of the rightmost node at each level

B. It does not check if root is null

C. It uses shift() which is inefficient

D. It pushes both left and right children in wrong order

Solution

1. Step 1: Analyze the condition for pushing node values

   The code pushes node.val when i === 0, which is the first node at each level (leftmost).

2. Step 2: Understand right side view requirement

   Right side view requires the rightmost node at each level, which is when i === size - 1.

3. Final Answer:

   It pushes the leftmost node instead of the rightmost node at each level -> Option A

4. Quick Check:

   Right side view needs last node per level, not first [OK]

Hint: Push node when i equals size-1 for right side view [OK]

Common Mistakes:

• Using i === 0 instead of i === size - 1
• Ignoring null root check
• Confusing order of child insertion

5. Given a binary tree where some nodes have only left children, how would the right side view look? Example tree:

    1
   /
  2
 / 
3

What is the right side view output for this tree?

hard

A. [3]

B. [1]

C. [1, 3]

D. [1, 2, 3]

Solution

1. Step 1: Visualize the tree and right side view

   Since all nodes are on the left, the right side view shows these nodes as they are the only visible ones.

2. Step 2: List nodes level-wise

   Level 1: 1, Level 2: 2, Level 3: 3. All are visible from right side as no right nodes block view.

3. Final Answer:

   [1, 2, 3] -> Option D

4. Quick Check:

   Right side view includes all nodes when only left children exist [OK]

Hint: If no right child, left child is visible from right side [OK]

Common Mistakes:

• Assuming right side view is empty if no right children
• Skipping left children in right side view
• Confusing right side view with right subtree nodes only