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: Accepted)

class Node
  attr_accessor :key, :left, :right, :parent
  def initialize key
    @key = key
  end
   
  def inorder
    left.inorder if left
    print " #{key}"
    right.inorder if right
  end
   
  def preorder
    print " #{key}"
    left.preorder if left
    right.preorder if right
  end
end
   
def insert
  z = Node.new($key.to_i)
  x = $root
  while x
    y = x
    x = z.key < x.key ? x.left : x.right
  end
   
  unless y
    $root = z
  else
    z.key < y.key ? y.left = z : y.right = z
    z.parent = y
  end
end
 
def find(x)
  y = $root
  while y
    return y if y.key == x
    z = y
    y = x < y.key ? y.left : y.right
  end
  return nil
end
 
def delete(x)
  y = find(x)
  z = position(y)
  if y.right.nil? || y.left.nil? then delete2(y,z)
  else
    p = y.right
    while p
      break if p.left == nil
      p = p.left
    end
    y.key = p.key
    z = position(p)
    delete2(p,z)
  end
end
 
def position(n)
  n.parent.right && n.parent.right.key == n.key
end
 
def delete2(y,z)
  if y.left.nil? && y.right.nil?
    z ? y.parent.right = nil : y.parent.left = nil
  elsif y.left.nil?
    if z 
      y.parent.right = y.right
      y.parent.right.parent = y.parent
    else
      y.parent.left = y.right
      y.parent.left.parent = y.parent
    end
  elsif y.right.nil?
    if z
      y.parent.right = y.left
      y.parent.right.parent = y.parent
    else
      y.parent.left = y.left
      y.parent.left.parent = y.parent
    end
  end
end
 
n = gets.to_i
n.times do
  command, $key = gets.split
  case command
  when "insert"
    insert
  when "find"
    if find($key.to_i)
      puts "yes"
    else
      puts "no"
    end
  when "delete"
    delete($key.to_i)
  else
    $root.inorder
    puts ""
    $root.preorder
    puts ""
  end
end