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Decode the Coding Interview in JavaScript: Real-World Examples

Binary Tree Right Side View

Explore how to determine the right side view of a binary tree by implementing a depth-first search in JavaScript. Learn to traverse the tree level by level from the rightmost node first and understand the time and space complexity of this approach.

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Description

You are given a binary tree T. Imagine you are standing to the right of it; you have to return the value of the nodes you can see from top to bottom.

You have to return the rightmost nodes on their respective levels in the form of an array.

Let’s discuss an example below:

Coding exercise

Files
index.js
TreeNode.js
Javascript (babel-node)
import {TreeNode} from './TreeNode.js'
function rightSideView(root){
  // write your code here
  
  return []
}
Binary tree right side view

Solution

We can use a depth-first search (DFS) to solve this problem. The intuition here is to traverse the tree level by level recursively, starting from the rightmost node for each recursive call.

Let’s review the implementation below:

Files
index.js
TreeNode.js
Javascript (babel-node)
import {TreeNode} from './TreeNode.js'
function rightSideView(root){
  if (root == null)
    return []
  var rightside = []
  
  function DFS(node, level){
    if (level == rightside.length)
      rightside.push(node.val)
    temp = [node.right, node.left]
    temp.forEach((child) => {
      if (child != null)
        DFS(child, level + 1)
    })
  }
          
  DFS(root, 0)
  return rightside
}
// Driver Code
var root = new TreeNode(1)
root.left = new TreeNode(2)
root.right = new TreeNode(3)
root.left.left = new TreeNode(4);
root.left.right = new TreeNode(5);
root.right.left = new TreeNode(6);
root.right.right = new TreeNode(7);
root.right.right.left = new TreeNode(8);
console.log(rightSideView(root))
Binary tree right side view

Complexity measures

Time Complexity Space
...