Problem:
Binary Search Tree II

Write a program which performs the following operations to a binary search tree $T$ by adding the find operation to A: Binary Search Tree I.

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

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

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

Input

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

10
insert 30
insert 88
insert 12
insert 1
insert 20
find 12
insert 17
insert 25
find 16
print

Sample Output 1

yes
no
1 12 17 20 25 30 88
30 12 1 20 17 25 88

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
  def initialize(val)
    @val = val
    @left = nil
    @right = nil
  end
  def insert(val)
    if val < @val
      if @left
        @left.insert(val)
      else
        @left = Node.new(val)
      end
    else
      if @right
        @right.insert(val)
      else
        @right = Node.new(val)
      end
    end
  end
  def print
    puts " #{inorder([]).join(" ")}"
    puts " #{preorder([]).join(" ")}"
  end
  def preorder(list)
    #root
    list << @val
    #left
    list = @left.preorder(list) if @left
    #right
    list = @right.preorder(list) if @right
    return list
  end
  def inorder(list)
    #left
    list = @left.inorder(list) if @left
    #root
    list << @val
    #right
    list = @right.inorder(list) if @right
    return list
  end
  def find(num)
    if @val == num then
      return "yes"
    elsif @val > num
      if @left
        return @left.find(num)
      else
        return "no"
      end
    else
      if @right
        return @right.find(num)
      else
        return "no"
      end
    end
  end
end

n = gets.to_i
tree = nil

n.times{
  command, num = gets.split
  case command;
  when "insert";
    if tree
      tree.insert(num.to_i)
    else
      tree = Node.new(num.to_i)
    end
  when "find";
    puts tree.find(num.to_i)
  when "print";
    tree.print
  end
}