Problem:
Binary Search Tree III

Write a program which performs the following operations to a binary search tree $T$ by adding delete operation to B: Binary Search Tree II.

insert $k$: Insert a node containing $k$ as key into $T$.

find $k$: Report whether $T$ has a node containing $k$.

delete $k$: Delete a node containing $k$.

print: Print the keys of the binary search tree by inorder tree walk and preorder tree walk respectively.

The operation delete $k$ for deleting a given node $z$ containing key $k$ from $T$ can be implemented by an algorithm which considers the following cases:

If $z$ has no children, we modify its parent $z.p$ to replace $z$ with NIL as its child (delete $z$).

If $z$ has only a single child, we "splice out" $z$ by making a new link between its child and its parent.

If $z$ has two children, we splice out $z$'s successor $y$ and replace $z$'s key with $y$'s key.

Input

In the first line, the number of operations $m$ is given. In the following $m$ lines, operations represented by insert $k$, find $k$, delete $k$ or print are given.

Output

For each find $k$ operation, print "yes" if $T$ has a node containing $k$, "no" if not.

In addition, for each print operation, print a list of keys obtained by inorder tree walk and preorder tree walk in a line respectively. Put a space character before each key

Constraints

The number of operations $\leq 500,000$

The number of print operations $\leq 10$.

$-2,000,000,000 \leq key \leq 2,000,000,000$

The height of the binary tree does not exceed 100 if you employ the above pseudo code.

The keys in the binary search tree are all different.

Sample Input 1

18
insert 8
insert 2
insert 3
insert 7
insert 22
insert 1
find 1
find 2
find 3
find 4
find 5
find 6
find 7
find 8
print
delete 3
delete 7
print

Sample Output 1

yes
yes
yes
no
no
no
yes
yes
1 2 3 7 8 22
8 2 1 3 7 22
1 2 8 22
8 2 1 22

Reference

Introduction to Algorithms, Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest, and Clifford Stein. The MIT Press.

Solution in Ruby (judge verdict: Wrong Answer)

class Node
  def initialize(parent, key)
    @key = key
    @parent = parent
    @left = nil
    @right = nil
  end

  attr_accessor :left, :right, :parent
  attr_reader :key

  def to_s
    "#{@key}"
  end
end

require 'set'
class BinarySearchTree
  def initialize
    @first = nil
    @visited_node = []
    @keys = Set.new
  end

  def insert(node)
    unless @first.nil?
      current = @first
      until current.nil?
        node.parent = current
        if node.key < current.key
          current = current.left
        else
          current = current.right
        end
      end
    end
    if @first.nil?
      @first = node
    elsif node.key < node.parent.key
      node.parent.left = node
    else
      node.parent.right = node
    end
    @keys << node.key
  end

  def find(key)
    if @keys.include? key
      puts 'yes'
    else
      puts 'no'
    end
  end

  def delete(key)
    unless @first.nil?
      current = @first
      until current.nil?
        if current.key == key
          delete1 current, key
          delete2 current, key
          delete3 current, key
          return
        end
        if key < current.key
          current = current.left
        else
          current = current.right
        end
      end
    end
  end

  def delete1(current, key)
    if current.left.nil? && current.right.nil?
      if current.parent.key > key
        current.parent.left = nil
      else
        current.parent.right = nil
      end
    end
  end

  def delete2(current, key)
    if current.left.nil? && !current.right.nil?
      if current.parent.key > key
        current.right.parent = current.parent
        current.parent.left = current.right
      else
        current.right.parent = current.parent
        current.parent.right = current.right
      end
    elsif !current.left.nil? && current.right.nil?
      if current.parent.key > key
        current.left.parent = current.parent
        current.parent.left = current.left
      else
        current.left.parent = current.parent
        current.parent.right = current.left
      end
    end

  end

  def delete3(current, key)
    unless current.left.nil? || current.right.nil?
      if current.parent.key > key
        current.right.parent = current.parent
        current.parent.left = current.right
      else
        current.right.parent = current.parent
        current.parent.right = current.right
      end
      delete1 current.right, key
      delete2 current.right, key
    end
  end

  def preorder_tree_walk(node)
    @visited_node << node.key
    return if node.nil?
    preorder_tree_walk node.left unless node.left.nil?
    preorder_tree_walk node.right unless node.right.nil?
  end

  def inorder_tree_walk(node)
    return if node.nil?
    inorder_tree_walk node.left unless node.left.nil?
    @visited_node << node.key
    inorder_tree_walk node.right unless node.right.nil?
  end

  def print
    inorder_tree_walk @first
    puts " #{@visited_node.join ' '}"
    @visited_node = []
    preorder_tree_walk @first
    puts " #{@visited_node.join ' '}"
    @visited_node = []
  end
end

n = gets.to_i
tree = BinarySearchTree.new

n.times do
  input = gets.split ' '
  case input[0]
  when 'insert'
    tree.insert Node.new nil, input[1].to_i
  when 'find'
    tree.find input[1].to_i
  when 'delete'
    tree.delete input[1].to_i
  when 'print'
    tree.print
  end
end