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: Time Limit Exceeded)

#include <stdio.h>
#define N 500000

typedef struct{
  int par;
  int left;
  int right;
  int key;
}node;

node T[N];
int root;

void Tree_insert(int,int);
void preorder(int);
void inorder(int);

int main(){

  int n,x,i,j,k=0;
  int find[N];
  char c[10];

  scanf("%d",&n);

  for(i=0; i<n; i++) T[i].left = T[i].right = T[i].key = -1;

  for(i=0; i<n; i++){

    scanf("%s",&c);

    if(c[0] == 'i'){
      scanf("%d",&x);
      Tree_insert(i,x);
    }


    if(c[0] == 'f'){

      scanf("%d",&x);

      for(j=0;j<i;j++){
	if(T[j].key == x){
	  find[k] = 1;
	  break;
	}
      }
      k++;
    }

    if(c[0] == 'p'){

      for(j=0;j<k;j++){
	if(find[j] == 1) printf("yes\n");
	else printf("no\n");

      }

      k = 0;

      inorder(0);
      printf("\n");
      preorder(0);
      printf("\n");

    }
  }

  for(j=0;j<k;j++){

    if(find[j] == 1) printf("yes\n");
    else printf("no\n");

  } 
  return 0;

}

void Tree_insert(int n,int x){

  int a,i;

  if(n == 0){

    T[n].par = -1; 
    T[n].key = x;
    root = 0;

  }

else{

    a = root;  

    while(1){

      if(T[a].key > x && T[a].left == -1){
	T[a].left = n;
	T[T[a].left].key = x;
	break;

      }

      if(T[a].key < x && T[a].right == -1){
	T[a].right = n;
	T[T[a].right].key = x;
	break;

      }

      if(T[a].key > x) a = T[a].left;
      else a = T[a].right;

    }			   

  }

}

void preorder(int n){

  printf(" %d",T[n].key);
 
  if(T[n].left != -1) preorder(T[n].left);

  if(T[n].right != -1) preorder(T[n].right);
}

void inorder(int n){
 
 if(T[n].left != -1) inorder(T[n].left);
  
printf(" %d",T[n].key);

  if(T[n].right != -1) inorder(T[n].right);
}