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 N 10

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



struct Node *sum,*num;

void jastin(int hum)
{

  struct Node *a=sum;
  struct Node *b=num;
  struct Node *c;

  c=(struct Node *)malloc(sizeof (struct Node));
  c->key=hum;
  c->left=num;
  c->right=num;

  while(a!=num)
    {
      b=a;
      if(c->key<a->key)a=a->left;
      else a=a->right;
    }
  c->parent=b;
  if(b==num)sum=c;
  else{
    if(c->key<b->key)b->left=c;
    else b->right=c;
  }
}
void ma(int key)
{
  struct  Node  *x=sum;

  while(x!=num){
      if(x->key >key)
	{
	  x=x->left;
	}
      else if(x->key < key)
	{
	  x=x->right;
	}
      else{
	printf("yes\n");
	return;
      }
    }
  
  printf("no\n");
}

void zin(struct Node *rum)
{
  if(rum==num)return;
  zin(rum->left);
  printf(" %d",rum->key);
  zin(rum->right);
}
void xu(struct Node *yum)
{
  if(yum==num)return;
  printf(" %d",yum->key);
  xu(yum->left);
  xu(yum->right);
}
int main()
{
  int i,j,k;
  char tum[N];

  scanf("%d",&i);
  
  for(j=0;j<i;j++)
    {
      scanf("%s",tum);
      if(tum[0]=='i')
	{
	  scanf("%d", &k);
	  jastin(k);
	}else if(tum[0]=='f'){
	  scanf("%d", &k);
	  ma(k);
	} else{ 
	zin(sum);
	printf("\n");
	xu(sum);
	printf("\n");
      }
      
      
    }
  return 0;
}