Problem:
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Binary Search Tree I

Search trees are data structures that support dynamic set operations including insert, search, delete and so on. Thus a search tree can be used both as a dictionary and as a priority queue.

Binary search tree is one of fundamental search trees. The keys in a binary search tree are always stored in such a way as to satisfy the following binary search tree property:

Let $x$ be a node in a binary search tree. If $y$ is a node in the left subtree of $x$, then $y.key \leq x.key$. If $y$ is a node in the right subtree of $x$, then $x.key \leq y.key$.

The following figure shows an example of the binary search tree.

For example, keys of nodes which belong to the left sub-tree of the node containing 80 are less than or equal to 80, and keys of nodes which belong to the right sub-tree are more than or equal to 80. The binary search tree property allows us to print out all the keys in the tree in sorted order by an inorder tree walk.

A binary search tree should be implemented in such a way that the binary search tree property continues to hold after modifications by insertions and deletions. A binary search tree can be represented by a linked data structure in which each node is an object. In addition to a key field and satellite data, each node contains fields left, right, and p that point to the nodes corresponding to its left child, its right child, and its parent, respectively.

To insert a new value $v$ into a binary search tree $T$, we can use the procedure insert as shown in the following pseudo code. The insert procedure is passed a node $z$ for which $z.key = v$, $z.left = NIL$, and $z.right = NIL$. The procedure modifies $T$ and some of the fields of $z$ in such a way that $z$ is inserted into an appropriate position in the tree.

1 insert(T, z)
2 y = NIL // parent of x
3 x = 'the root of T'
4 while x ≠ NIL
5 y = x // set the parent
6 if z.key < x.key
7 x = x.left // move to the left child
8 else
9 x = x.right // move to the right child
10 z.p = y
11
12 if y == NIL // T is empty
13 'the root of T' = z
14 else if z.key < y.key
15 y.left = z // z is the left child of y
16 else
17 y.right = z // z is the right child of y

Write a program which performs the following operations to a binary search tree $T$.

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

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

You should use the above pseudo code to implement the insert operation. $T$ is empty at the initial state.

Input

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

Output

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

8
insert 30
insert 88
insert 12
insert 1
insert 20
insert 17
insert 25
print

Sample Output 1

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 Scala (judge verdict: Memory Limit Exceeded)

import scala.io.StdIn.{readInt, readLine}
import scala.collection.mutable.ArrayBuffer

object Main {
  val tree0 = new Array[Int](500001)
  val tree1 = new Array[Int](500001)
  val tree2 = new Array[Int](500001)
  val a = new ArrayBuffer[Int]

  def insert(i:Int, j:Int):Unit =
    if(tree0(i) > tree0(j)) {
      if(tree1(i) == -1) tree1(i) = j
      else insert(tree1(i), j)
    } else {
      if(tree2(i) == -1) tree2(i) = j
      else insert(tree2(i), j)
    }

  def inOrder():Unit = {
    var stack:List[Int] = Nil
    var curr = 0
    var flg = true

    while(flg) {
      while(curr != -1) {
        stack = curr::stack
        curr = tree1(curr)
      }
      if(curr == -1 && stack != Nil) {
        val item = stack.head
        stack = stack.tail
        a.append(tree0(item))
        curr = tree2(item)
      } else {
        flg = false
      }
    }
  }

  def scan(list:List[Int]):Unit = {
    list match {
      case Nil => ()
      case hd::tl => {
        a.append(tree0(hd))
        val hd2 =
          if(tree2(hd) != -1 && tree1(hd) != -1) tree1(hd)::tree2(hd)::tl
          else if(tree2(hd) != -1 && tree1(hd) == -1) tree2(hd)::tl
          else if(tree2(hd) == -1 && tree1(hd) != -1) tree1(hd)::tl
          else tl
        scan(hd2)
      }
    }
  }

  def preOrder(i:Int):Unit =
    if(i == -1) return
    else scan(i::Nil)

  def main(args:Array[String]) = {
    val n = readInt
    var j = 0

    for(i<-0 until n) {
      val s = readLine.split(" ")
      s(0) match {
        case "insert" => {
          tree0(j) = s(1).toInt
          tree1(j) = -1
          tree2(j) = -1
          if(j!=0) insert(0, j)
          j+=1
        }
        case _ => {
          a.clear
          inOrder();
          print(" ");println(a.mkString(" "))
          a.clear
          preOrder(0)
          print(" ");println(a.mkString(" "))
        }
      }
    }
  }
}