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 nd{
  int key;
  struct nd *parent,*right,*left;
} ;


struct nd *root,*nil;

struct nd *find(struct nd *n,int z){
  while(n != nil && z != n->key){
    if(z < n->key){
      n = n->left;
    }
    else{
      n = n->right;
    }
  }
  return n;
}

  

void insert(int z){
  struct nd *x,*y;
  struct nd *n;
  y=nil;
  x=root;
  
  n=(struct nd *)malloc(sizeof(struct nd));
  n->key=z;
  n->left=nil;
  n->right=nil;
  
  while(x!=nil){
    y=x;
    if(n->key<x->key){
      x=x->left;
    }
    else{
      x=x->right;
    }
  }
  n->parent=y;
  if(y==nil){
    root=n;
  }
  else{
    if(n->key<y->key){
      y->left=n;
    }
    else {
      y->right=n;
    }
  }
}

void pri1(struct nd *n){
  if(n==nil){
    return ;
  }
  pri1(n->left);
  printf(" %d",n->key);
  pri1(n->right);
}

void pri2(struct nd *n){
  if(n==nil){
    return ;
  }
  printf(" %d",n->key);
  pri2(n->left);
  pri2(n->right);
}


    
int main(){
  int kazu,key;
  char cmd[20];
  int i,k,j;
  scanf("%d",&kazu);
  for(i=0;i<kazu;i++){
    scanf("%s",cmd);
    scanf("%d",&key);
    if(cmd[0]=='i'){
      insert(key);
    }
    else if(cmd[0]=='p'){
      pri1(root);
      printf("\n");
      pri2(root);
      printf("\n");
    }
    else if(cmd[0]=='f'){
      struct nd *tai=find(root,key);
      if(tai!=nil){
	printf("yes\n");
      }
      else{
	printf("no\n");
      }
    }
  }
  
  return 0;
}