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<string.h>
#include<stdlib.h>

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

typedef struct Node Node;

Node *NIL;
Node *root;

Node *find(Node *u,int k);
void insert(int k);
void inorder(Node *u);
void preorder(Node *u);



int main(void)
{
  int n,x;
  int i;
  char com[20];

  scanf("%d",&n);

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

      if(com[0]=='f')
	{
	  scanf("%d",&x);
      
	  Node *t=find(root,x);
      
	  if(t!=NIL)
	    {
	      puts("yes");
	    }

	  else
	    {
	      puts("no");
	    }
      
	}
    
      else if(strcmp(com,"insert") == 0)
	{
	  scanf("%d",&x);
      
	  insert(x);    
	}
    
      else if(strcmp(com,"print") == 0)
	{
	  inorder(root);
      
	  putchar('\n');
      
	  preorder(root);
      
	  putchar('\n');
	}
    
    }
  
  return 0;
}

Node * find(Node *u,int k)
{
  while(u!=NIL&&k!=u->key)
    {
      u=(u->key<k)?u->right:u->left;
    }

  return u;
}

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

  z=(Node *)malloc(sizeof(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);
}