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>
#define MAX 500001
#define NIL -1
typedef struct node{
  int left;
  int right;
  int key;
  int p;
  int leftid;
  int rightid;

}Node;
Node node[MAX];
int root=NIL;
void inorder(int);
void preorder(int);
int search(int s,int r){
  if(r==NIL){
    return 0;
  }
  if(node[r].key==s){
    return 1;
  }
  else {
    if(node[r].key>s &&node[r].left!=NIL){
    return  search(s,node[r].leftid);
    }
    if(node[r].key<s&&node[r].right!=NIL){
      return search(s,node[r].rightid);
    }
    return 0;
  }

}
void addTree(int n,int num){
  int tmp;
  //rootが空の場合
  if(node[n].key==root){
    root=n;
    node[root].key=num;
    
  }
  else{
  tmp=root;

  while(1){
    if(node[tmp].key>num){
      if(node[tmp].left!=NIL){
	tmp=node[tmp].leftid;
      }
      else{
	node[tmp].left=num;
	node[tmp].leftid=n+1;
	node[n+1].key=num;
	
	break;
      }
    }
    else{
      if(node[tmp].right!=NIL){
	tmp=node[tmp].rightid;
	
      }
      else{
	node[tmp].right=num;
	node[n+1].key=num;
	node[tmp].rightid=n+1;
	break;
      }
    }
  }
  
  node[n].p=tmp;
  
  }
}
int main(){
  int num,i,n,l=0,tmp,s,flag=0;
  char name[100];
  int judge;
  scanf("%d",&n);
 for(i=0;i<MAX;i++){//初期化
    node[i].p=NIL;
    node[i].left=NIL;
    node[i].right=NIL;
    node[i].key=NIL;
    judge=0;
  }


 for(i=0;i<n;i++){
    scanf("%s",name);
    if(name[0]=='i'){
      scanf("%d",&num);
      addTree(l,num);
      l++;
      }
      
   else if(name[0]=='p'){

    inorder(root);
   printf("\n");
   preorder(root);
   printf("\n");
   continue;
    }
   else if(name[0]=='f'){
     scanf("%d",&s);
     judge= search(s,root);
    
    if(judge==0){

      
    printf("no\n");
    }
    else{
    
      printf("yes\n");
    }
    judge=0;
    continue;
   }
  }
 
  return 0;
}
void inorder(int t){
  if(node[t].left!=NIL){
    inorder(node[t].leftid);
  }
    printf(" %d",node[t].key);
  if(node[t].right!=NIL){
    inorder(node[t].rightid);
  }
}
void preorder(int t){
 
      printf(" %d",node[t].key);
  if(node[t].left!=NIL){
    preorder(node[t].leftid);
  }
  if(node[t].right!=NIL){
    preorder(node[t].rightid);
  }
}