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

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->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){
    root = z;
  }else{
    if(z->key < y->key){
      y->l = z;
    }else{
      y->r = z;
    }
  }
}

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);
}

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

int main(){
  int n, i, x;
  char com[20];
  scanf("%d", &n);

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

  return 0;
}