Why Right Side View of Binary Tree in DSA Javascript?

The Big Idea

What if you could instantly see only the parts of a tree visible from one side without guessing?

The Scenario

Imagine you are standing on the right side of a tall tree and want to see only the branches visible from your side. If you try to list all branches manually from the top to bottom, it becomes confusing and hard to know which ones are actually visible.

The Problem

Manually checking each branch from the right side is slow and error-prone because you might miss branches hidden behind others or waste time looking at branches that are not visible from the right side.

The Solution

The "Right Side View of Binary Tree" concept helps by automatically finding and showing only the branches visible from the right side, ignoring hidden ones. It simplifies the problem by scanning the tree level by level and picking the rightmost branch at each level.

Before vs After

✗ Before

function rightSideView(root) {
  // Manually check each node and guess visibility
  // This is complicated and error-prone
}

✓ After

function rightSideView(root) {
  if (!root) return [];
  let result = [];
  let queue = [root];
  while(queue.length) {
    let size = queue.length;
    for(let i = 0; i < size; i++) {
      let 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;
}

What It Enables

This concept enables you to quickly see the outline of a tree from the right side, which helps in visualizing and understanding tree structures clearly.

Real Life Example

In a city skyline, you want to see the buildings visible from the right side street. This concept helps you list only those buildings that are not hidden behind others when viewed from that side.

Key Takeaways

Manually finding right side view is confusing and slow.

Right Side View uses level order traversal to pick rightmost nodes.

It helps visualize tree structure from one side easily.

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