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>

struct Node{
  int key;
  struct Node *r,*l,*p;
};

 struct Node *B,*NIL;

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

   z = (struct Node *)malloc(sizeof(struct Node));
   z->key = k;
   z->l = NIL;
   z->r = NIL;

   while(x!=NIL){
     y = x;
     
     if(z->key < x->key){
       x = x->l;
     }
     else x = x->r;
   }
   
   z->p = y;
   if(y == NIL ){
     B=z;
   }
   else if(z->key < y->key){
       y->l = z;
     }
   else y->r = z;
}

struct Node* find(int k,struct Node *u){
  while( u != NIL && k != u->key ){
    if(k < u->key) u = u->l;
    else u = u->r;
  }
  return u;
}


void inorder(struct Node *u){
  if(u==NIL) return;
  inorder(u->l);
  printf(" %d",u->key);
  inorder(u->r);
}

void preorder(struct Node *u){
  if(u == NIL )return;
  printf(" %d",u->key);
  preorder(u->l);
  preorder(u->r);
}

int main(){
  
  int i,j,x,y;
  char G[8];
  struct Node *s;
  scanf("%d",&x);
  
  for(i=0;i<x;i++){
    scanf("%s",&G);
    
    if(G[0] == 'i'){
      scanf("%d",&y);
      insert(y);
    }
    
    else if(G[0] == 'p'){
      inorder(B);
      printf("\n");
      preorder(B);
      printf("\n");
    }

    else if(G[0]=='f'){
      scanf("%d",&y);
      s = find(y,B);
      if(s != NIL) printf("yes\n");
      else if(s == NIL)printf("no\n");
    }
  }
  return 0;
}