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>
#define TRUE 1
#define FALSE 0
struct node{
  int key;
  struct node* right;
  struct node* left;
  struct node* parent;
};
struct node *root,*NIL;

void insert(int data){
  struct node *p = NIL;
  struct node *x = root;
  struct node *new;

  new =(struct node*)malloc(sizeof(struct node));
  new->key = data;
  new->left = NIL;
  new->right = NIL;

  //x が root ではない。つまり、値が代入されていない。または、全て出力された場合。
  while(x != NIL){
    p = x;
    if(new->key < x->key){
      x  = x->left;
    }else{
      x = x->right;
    }
  }
  new->parent = p;
  if(p == NIL){
    root = new;
  }else{
    if(new->key < p->key){
      p->left = new;
    }else{
      p->right = new;
    }
  }
}

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);
}

void print(){
  inorder(root);
  printf("\n");
  preorder(root);
  printf("\n");
  return;
}
int find(int data,struct node* temp){
  //data != NILすなわち値がない行き詰まりの時、かつ、keyの値と同じ値が見つからない間
  while(temp != NIL && data != temp->key){
    if(temp->key > data)temp = temp->left;
    else temp = temp->right;
  }
  if(temp == NIL)return FALSE;
  else return TRUE;
}

int main(){
  int n,i,x,temp,ans;
  char com[10];
  scanf("%d",&n);
  for(i=0;i<n;i++){
    scanf("%s",com);
    if(strcmp("insert",com) == 0){
      scanf("%d",&temp);
      insert(temp);
    }else if(strcmp("print",com) == 0){
      print();
    }else if(strcmp("find",com) == 0){
      scanf("%d",&temp);
      ans = find(temp,root);
      if(ans == TRUE)printf("yes\n");
      else printf("no\n");
      //rootからスタートして、tempの値があるかどうかを探索する。
    }else{
      printf("command error!\n");
    }
  }
  return 0;
}