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 C (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#define NIL NULL


struct node{
  struct node *right;
  struct node *left;
  struct node *parent;
  int key;
};
typedef struct node * Node;

Node root;

void insert(int k){
  Node y = NIL;
  Node x = root;
  Node z;

  z = malloc(sizeof(struct node));
  
  z->key = k;
  z->left = NIL;
  z->right = NIL;

  while( x != NIL){
    y = x;
    if(z->key < x->key){
      x =x->left;
    }else{
      x = x->right;
    }
  }
  z->parent = y;
  if(y == NIL){
    root = z;
  }else {
    if(z->key < y->key){
      y->left = z;
    }else{
      y->right = z;
    }
  }
}

void inorder(Node u){
  if(u == NIL) return;
  inorder(u->left);
  printf(" %d",u->key );
  inorder(u->right);
}

void preorder(Node u){
  if(u == NIL) return;
  printf(" %d",u->key );
  preorder(u->left);
  preorder(u->right);
}

Node treeSearch(Node u, int k){
  while( u != NIL && k != u->key){
    if(k < u->key) u = u->left;
    else u = u->right;
  }
  return u;
}



int main(){
  int n/*最初の入力数*/, i, x;
  char com[20];/*insert or print*/

  scanf("%d", &n);/*最初の入力数*/

  for ( i = 0; i < n; i++ ){
    scanf("%s", com);
    
    if ( com[0] == 'i' ){

      scanf("%d", &x);/* insert */
      insert(x);

    } else if ( com[0] == 'p' ){

      inorder(root);
      printf("\n");

      preorder(root);
      printf("\n");

    }else  if ( com[0] == 'f' ){
      scanf("%d", &x);
      Node t = treeSearch(root, x);
      if ( t != NIL ) printf("yes\n");
      else printf("no\n");
  }
}

  return 0;
}