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>
#include <string.h>

struct node{
  int key;
  struct node *p;
  struct node *l;
  struct node *r;
};
struct node *head;


void pre(struct node *u){
  if(u->key == -1){
    return;
  }
  printf(" %d", u->key);
  pre(u->l);
  pre(u->r);
}

void in(struct node *u){
  if(u->key == -1){
    return;
  }
  in(u->l);
  printf(" %d", u->key);
  in(u->r);
}

void insert(int u){
  struct node *z;
  z = (struct node *)malloc(sizeof(struct node));
  z->key = u;
  z->l = (struct node *)malloc(sizeof(struct node));
  z->r = (struct node *)malloc(sizeof(struct node));
  z->l->key = -1;
  z->r->key = -1;
  struct node *y;
  struct node *x;
  x = head;
  y = head->p;

  while(x->key != -1){
    y = x;
    if(u < x->key){
      x = x->l;
    }else{
      x = x->r;
    }
  }
    
  z->p = y;
  if(y->key == -1){
    head = z;
  }else if(z->key < y->key){
    y->l = z;
  }else{
    y->r = z;
  }
}

int findr(int k, struct node *z){
  if(z->key == -1){
    return 0;
  }else if(z->key == k){
    return 1;
  }
  if(k < z->key){
    return findr(k, z->l);
  }else{
    return findr(k, z->r);
  }
}

int find(int k){
  int l = findr(k, head);
  return l;
}

void print(void){
  in(head);
  printf("\n");
  pre(head);
  printf("\n");
}

int main(void){
  int i, n, x;
  head = (struct node *)malloc(sizeof(struct node));
  head->key = -1;
  head->l = (struct node *)malloc(sizeof(struct node));
  head->r = (struct node *)malloc(sizeof(struct node));
  head->p = (struct node *)malloc(sizeof(struct node));
  head->p->key = -1;
  head->l->key = -1;
  head->r->key = -1;
  char s[16];
  scanf("%d", &n);
  for(i = 0; i < n; i++){
    scanf("%s", s);
    if(!strcmp(s, "insert")){
      scanf("%d", &x);
      insert(x);
    }else if(!strcmp(s, "print")){
      print();
    }else if(!strcmp(s, "find")){
      scanf("%d", &x);
      if(find(x)){
	printf("yes\n");
      }else{
	printf("no\n");
      }
    }
  }
  return 0;
}