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: Runtime Error)

#include <stdio.h>
#include <string.h>
#define N 500000
int count;

struct node
{
  int key;
  int pare;
  int left;
  int right;
};

void insert(struct node bi[],int z)
{
  int x,y;
  int num=0;

  y=-1;
  x=bi[0].key;
//  printf("while start\n");
  while(1)
  {
    if(x==-1) break;
    y = num;
    if(z<x)
    {
      num = bi[num].left;
      if(num!=-1) x = bi[num].key;
      else
      {
        x=-1;
      }
    }
    else
    {
      num = bi[num].right;
      if(num!=-1) x = bi[num].key;
      else
      {
        x=-1;
      }
    }
  }
//  printf("while end\n");
  bi[count].pare=y;
  bi[count].key=z;

  if(y==-1)
  {
    bi[0].key=z;
  }
  else if(z<bi[y].key)
  {
    bi[y].left = count;
  }
  else
  {
    bi[y].right = count;
  }

  count++;
}

void pre(struct node bi[],int now)
{
  printf(" %d",bi[now].key);
  if(bi[now].left!=-1)  pre(bi,bi[now].left);
  if(bi[now].right!=-1)  pre(bi,bi[now].right);
}

void in(struct node bi[],int now)
{
  if(bi[now].left!=-1) in(bi,bi[now].left);
  printf(" %d",bi[now].key);
  if(bi[now].right!=-1) in(bi,bi[now].right);
}


void print(struct node bi[])
{
  in(bi,0);
  printf("\n");
  pre(bi,0);
  printf("\n");
}

void find(int key,struct node bi[])
{
  int i=0;
  while(1)
  {
    if(bi[i].pare==-1) break;
    if(bi[i].key==key)
    {
      printf("yes\n");
      return;
    }
    i = bi[i].pare;
  }
  while(1)
  {
    if(bi[i].key<key)
    {
      i=bi[i].right;
      if(bi[i].key==-1)
      {
        printf("no\n");
        return;
      }
    }
    else
    {
      i=bi[i].left;
      if(bi[i].key==-1)
      {
        printf("no\n");
        return;
      }
    }
    if(bi[i].key==key)
    {
      printf("yes\n");
      return;
    }
  }
}

int main(void)
{
  int n,i;
  int num;
  char comm[10];
  struct node bi[N];
  count=0;

  for(i=0;i<N;i++)
  {
    bi[i].key = -1;
    bi[i].pare = -1;
    bi[i].left = -1;
    bi[i].right = -1;
  }
//  printf("set ok\n");

  scanf("%d",&n);

  for(i=0;i<n;i++)
  {
//    printf("comm:");
    scanf("%s",comm);
//    printf("comm:%s\n",comm);
    if(strcmp(comm,"insert")==0)
    {
//      printf("scan:");
      scanf("%d",&num);
      insert(bi,num);
    }
    else if(strcmp(comm,"find")==0)
    {
      scanf("%d",&num);
      find(num,bi);
    }
    else
    {
//      printf("print\n");
      print(bi);
    }
  }
  return 0;
}