0606-construct-string-from-binary-tree.rb

# frozen_string_literal: true

# 606. Construct String from Binary Tree
# Easy
# https://leetcode.com/problems/construct-string-from-binary-tree

=begin
Given the root of a binary tree, construct a string consisting of parenthesis and integers from a binary tree with the preorder traversal way, and return it.
Omit all the empty parenthesis pairs that do not affect the one-to-one mapping relationship between the string and the original binary tree.

Example 1:
Input: root = [1,2,3,4]
Output: "1(2(4))(3)"
Explanation: Originally, it needs to be "1(2(4)())(3()())", but you need to omit all the unnecessary empty parenthesis pairs. And it will be "1(2(4))(3)"

Example 2:
Input: root = [1,2,3,null,4]
Output: "1(2()(4))(3)"
Explanation: Almost the same as the first example, except we cannot omit the first parenthesis pair to break the one-to-one mapping relationship between the input and the output.

Constraints:
The number of nodes in the tree is in the range [1, 104].
-1000 <= Node.val <= 1000
=end

# Definition for a binary tree node.
# class TreeNode
#     attr_accessor :val, :left, :right
#     def initialize(val = 0, left = nil, right = nil)
#         @val = val
#         @left = left
#         @right = right
#     end
# end
# @param {TreeNode} root
# @return {String}
def tree2str(root)
  return "" if root.nil?

  string = root.val.to_s
  string += "(#{tree2str(root.left)})" if root.left || root.right
  string += "(#{tree2str(root.right)})" if root.right

  string
end