0173-binary-search-tree-iterato.rb

# frozen_string_literal: true

# 173. Binary Search Tree Iterator
# https://leetcode.com/problems/binary-search-tree-iterator
# Medium

=begin

Implement the BSTIterator class that represents an iterator over the in-order traversal of a binary search tree (BST):

* BSTIterator(TreeNode root) Initializes an object of the BSTIterator class. The root of the BST is given as part of the constructor. The pointer should be initialized to a non-existent number smaller than any element in the BST.
* boolean hasNext() Returns true if there exists a number in the traversal to the right of the pointer, otherwise returns false.
* int next() Moves the pointer to the right, then returns the number at the pointer.
Notice that by initializing the pointer to a non-existent smallest number, the first call to next() will return the smallest element in the BST.

You may assume that next() calls will always be valid. That is, there will be at least a next number in the in-order traversal when next() is called.

### Example 1:
Input
["BSTIterator", "next", "next", "hasNext", "next", "hasNext", "next", "hasNext", "next", "hasNext"]
[[[7, 3, 15, null, null, 9, 20]], [], [], [], [], [], [], [], [], []]
Output
[null, 3, 7, true, 9, true, 15, true, 20, false]

Explanation
BSTIterator bSTIterator = new BSTIterator([7, 3, 15, null, null, 9, 20]);
bSTIterator.next();    // return 3
bSTIterator.next();    // return 7
bSTIterator.hasNext(); // return True
bSTIterator.next();    // return 9
bSTIterator.hasNext(); // return True
bSTIterator.next();    // return 15
bSTIterator.hasNext(); // return True
bSTIterator.next();    // return 20
bSTIterator.hasNext(); // return False

### Constraints:
* The number of nodes in the tree is in the range [1, 105].
* 0 <= Node.val <= 106
* At most 105 calls will be made to hasNext, and next.

=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
class BSTIterator
=begin
    :type root: TreeNode
=end
  def initialize(root)
    @stack = [root]
  end

=begin
    :rtype: Integer
=end
  def next
    find_next
  end


=begin
    :rtype: Boolean
=end
  def has_next
    !@stack.empty?
  end

  private
    def find_next
      current = @stack.pop
      return current if current.is_a?(Integer)
      @stack += [current.right, current.val, current.left].compact
      find_next
    end
end

# Your BSTIterator object will be instantiated and called as such:
# obj = BSTIterator.new(root)
# param_1 = obj.next()
# param_2 = obj.has_next()