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, *left, *parent;
};

struct Node *root, *NIL;

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

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

}

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



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

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

	
int main(void){

  int n, i, x;
  char str[32];
 
  scanf("%d",&n);
  
  for(i=0;i<n;i++){
    scanf("%s",str);
    if(str[0] == 'f'){
      scanf("%d", &x);
      struct Node *t = find(root, x);
      if(t != NIL){
	printf("yes\n");
      }else{
	printf("no\n");
      }
    }else if(strcmp(str,"insert") == 0){
      scanf("%d", &x);
      insert(x);
    }else if(strcmp(str,"print") == 0){
      inorder(root);
      printf("\n");
      preorder(root);
      printf("\n");
    }
  }

return 0;


}