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 *right;
  struct node *left;
  struct node *parent;
  int key;
};
typedef struct node * Node;
#define NIL NULL

Node root;

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

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

  z = 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;
  } 
}

void inorder(Node u){

  if(u->left != NIL){
    inorder(u->left);
  }

  printf(" %d",u->key);

  if(u->right != NIL){
    inorder(u->right);
  }

}

void preorder(Node u){

  printf(" %d",u->key);

  if(u->left != NIL){
    preorder(u->left);
  }

  if(u->right != NIL){
    preorder(u->right);
  }

}

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

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

  return 0;
}