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Decode the Coding Interview in Ruby: Real-World Examples

Binary Tree Right Side View

Explore how to extract the rightmost visible nodes from each level of a binary tree using depth-first search. Understand the approach and complexity to help you solve tree traversal problems confidently.

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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

Ruby
require './TreeNode.rb'
def right_side_view(root)
  # write your code here
end
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:

Ruby
require './TreeNode.rb'
def right_side_view(root) 
  if root == nil 
      return []
  end
  rightside = []
  
  def DFS(node, level, rightside) 
      if level == rightside.length() 
          rightside.push(node.val)
      end
      [node.right, node.left].each do |child| 
          if child 
              DFS(child, level + 1, rightside)
          end
      end
  end
  
  DFS(root, 0, rightside)
  return rightside
end
# Driver Code
root = TreeNode.new(1)
root.left = TreeNode.new(2)
root.right = TreeNode.new(3)
root.left.left = TreeNode.new(4);
root.left.right = TreeNode.new(5);
root.right.left = TreeNode.new(6);
root.right.right = TreeNode.new(7);
root.right.right.left = TreeNode.new(8);
p(right_side_view(root))
Binary tree right side view

Complexity measures

Time Complexity Space
...