Right Side View of Binary Tree in DSA Typescript - Build from Scratch

Right Side View of Binary Tree

📖 Scenario: You are working with a binary tree, which is like a family tree but each person has at most two children: a left child and a right child. You want to see the tree from the right side, so you only see the rightmost nodes at each level.

🎯 Goal: Build a program that finds the right side view of a binary tree. This means you will list the values of the nodes that are visible when looking at the tree from the right side, level by level.

📋 What You'll Learn

Create a binary tree using a TreeNode class with val, left, and right properties

Create a variable called root that represents the root of the binary tree with exact values

Create a variable called rightSideView to store the visible nodes from the right side

Use a breadth-first search (BFS) approach with a queue to traverse the tree level by level

At each level, add the value of the rightmost node to rightSideView

Print the rightSideView array at the end

💡 Why This Matters

🌍 Real World

Right side view of a binary tree helps in visualizing hierarchical data from a specific angle, useful in UI trees, organizational charts, and decision trees.

💼 Career

Understanding tree traversal and views is important for software engineers working with data structures, algorithms, and system design.

1

Create the Binary Tree Structure

Create a class called TreeNode with a constructor that takes a number val and optional left and right nodes. Then create a variable called root that builds this exact tree:

1
/ \
2 3
\ \
5 4

DSA Typescript

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

// Your code here

Hint

Think of each node as a person with a value and two children. Use the constructor to connect them exactly as shown.

2

Set Up the Result Array

Create a variable called rightSideView and set it to an empty array of numbers. This will hold the values of the nodes visible from the right side.

DSA Typescript

class TreeNode {
  val: number;
  left: TreeNode | null;
  right: TreeNode | null;
  constructor(val: number, left: TreeNode | null = null, right: TreeNode | null = 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))
);

// Your code here

Hint

This array will collect the rightmost node values from each level.

3

Traverse the Tree Level by Level

Create a queue array called queue and add root to it. Use a while loop that runs while queue.length > 0. Inside the loop, get the number of nodes at the current level with levelLength = queue.length. Then use a for loop with variable i from 0 to levelLength - 1. Remove the first node from queue and call it node. If node.left exists, add it to queue. If node.right exists, add it to queue. If i === levelLength - 1, add node.val to rightSideView.

DSA Typescript

class TreeNode {
  val: number;
  left: TreeNode | null;
  right: TreeNode | null;
  constructor(val: number, left: TreeNode | null = null, right: TreeNode | null = 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))
);

const rightSideView: number[] = [];

// Your code here

Hint

Use a queue to visit nodes level by level. Add children to the queue. The last node at each level is the rightmost one.

4

Print the Right Side View

Write a console.log statement to print the rightSideView array.

DSA Typescript

class TreeNode {
  val: number;
  left: TreeNode | null;
  right: TreeNode | null;
  constructor(val: number, left: TreeNode | null = null, right: TreeNode | null = 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))
);

const rightSideView: number[] = [];

const queue: (TreeNode | null)[] = [root];

while (queue.length > 0) {
  const levelLength = queue.length;
  for (let i = 0; i < levelLength; i++) {
    const node = queue.shift()!;
    if (node.left) queue.push(node.left);
    if (node.right) queue.push(node.right);
    if (i === levelLength - 1) {
      rightSideView.push(node.val);
    }
  }
}

// Your code here

Hint

Print the array that holds the right side view values.