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 <string.h>
#define N 500000
int count;
 
struct node
{
  int key;
  int pare;
  int l;
  int r;
};
 
struct node bi[N];
 
void insert(int z)
{
  int x,y;
  int num = 0;
 
  y = -1;
  x = bi[0].key;

  while(1)
  {
    if(x==-1)break;

    y = num;

    if(z<x)
    {
      num = bi[num].l;
      if(num!=-1)x = bi[num].key;
      else
      {
        x = -1;
      }
    }
    else
    {
      num = bi[num].r;
      if(num != -1)x = bi[num].key;
      else
      {
        x = -1;
      }
    }
  }

  bi[count].pare=y;
  bi[count].key=z;
 
  if(y==-1)
  {
    bi[0].key=z;
  }
  else if(z < bi[y].key)
  {
    bi[y].l = count;
  }
  else
  {
    bi[y].r = count;
  }
 
  count++;
}
 
void pre(int now)
{
  printf(" %d",bi[now].key);
  if(bi[now].l != -1)  pre(bi[now].l);
  if(bi[now].r != -1)  pre(bi[now].r);
}
 
void in(int now)
{
  if(bi[now].l != -1)in(bi[now].l);
  printf(" %d",bi[now].key);
  if(bi[now].r != -1)in(bi[now].r);
}
 
 
void print()
{
  in(0);
  printf("\n");
  pre(0);
  printf("\n");
}
 
void find(int key)
{
  int i = 0;
  while(1)
  {
    if(bi[i].key == key)
    {
      printf("yes\n");
      return;
    }
    if(bi[i].pare == -1)break;
    i = bi[i].pare;
  }

  while(1)
  {
    if(bi[i].key < key)
    {
      i = bi[i].r;
      if(i == -1)
      {
        printf("no\n");
        return;
      }
    }
    else
    {
      i = bi[i].l;
      if(i == -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];
  count=0;
 
  for(i = 0;i < N;i++)
  {
    bi[i].key = -1;
    bi[i].pare = -1;
    bi[i].l = -1;
    bi[i].r = -1;
  }

 
  scanf("%d",&n);
 
  for(i = 0;i < n;i++)
  {

    scanf("%s",comm);

    if(strcmp(comm,"insert") == 0)
    {

      scanf("%d",&num);
      insert(num);
    }
    else if(strcmp(comm,"find") == 0)
    {
      scanf("%d",&num);
      find(num);
    }
    else
    {
      print();
    }
  }
  return 0;
}