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 JAVA (judge verdict: Accepted)

import java.io.*;

public class Main {

  static int n, pos, ansInp;
  static int[] pre, inp, post;

  
  public static void main(String[] args) throws Exception {
    
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    
    Tree t = new Tree();

    String[] in = br.readLine().split(" ");
    n = Integer.parseInt(in[0]);

    for(int i = 0; i < n; i++){
      in = br.readLine().split(" ");
      if(in[0].equals("insert")){
        t.insert(new Node(Integer.parseInt(in[1])));
      }else if(in[0].equals("print")){
        t.print();
      }else if(in[0].equals("find")){
        System.out.println(t.find(Integer.parseInt(in[1])) ? "yes" : "no");
      }
    }

    br.close();

  }
}


class Node {
  int v;
  Node p, l, r;

  Node(int v){
    this.v = v;
  }
}

class Tree {

  Node root;

  public Node insert(Node node){
    Node par = null;
    Node now = root;

    while(now != null){
      par = now;
      if(node.v > now.v) now = now.r;
      else now = now.l;
    }
    node.p = par;

    if(par == null) {
      this.root = node;
    }else if(node.v < par.v) {
      par.l = node;
    }else {
      par.r = node;
    }
    return node;
  }

  void print(){
    inorderPrint(this.root);
    System.out.println();
    preorderPrint(this.root);
    System.out.println();
  }

  void preorderPrint(Node node){
    System.out.print(' ');
    System.out.print(node.v);
    if(node.l != null)
      preorderPrint(node.l);
    if(node.r != null)
      preorderPrint(node.r);
    return;
  }

  void inorderPrint(Node node){
    if(node.l != null)
      inorderPrint(node.l);
    System.out.print(' ');
    System.out.print(node.v);
    if(node.r != null)
      inorderPrint(node.r);
    return;
  }

  boolean find(int x){
    boolean ret = false;
    Node now = root;
    while(now != null){
      if(now.v == x){
        ret = true;
        break;
      }else if(now.v > x){
        now = now.l;
      }else{
        now = now.r;
      }
    }
    return ret;
  }
}