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>

struct Node{
  int key;
  struct Node *right;
  struct Node *left;
  struct Node *parent;
};

struct Node *root;
struct Node *NIL;


struct Node *find(struct 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){

  struct Node *y = NIL;
  struct Node *x = root;
  struct Node *z;

  z = (struct Node *)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(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);
}

int main(){

  int n;
  int i,j;
  int x;
  
int count=0;
int cnt[10];
int tmp=0;

  struct Node *t;
  char *com;
  char *comin = "insert";
  char *compr = "print";

  com = (char*)malloc(7*sizeof(char));  

  scanf("%d", &n);

  for(i=0; i<n; i++){
    scanf("%s", com);
    if(com[0] == 'f'){
      scanf("%d", &x);
      t = find(root, x);
      if(t != NIL){
	printf("yes\n");
      }
      else{
	printf("no\n");
      }
    }
      else if(strncmp(com,comin,6)==0){
	scanf("%d",&x);
	insert(x);
      }
      else if(strncmp(com,compr,5)==0){

	inorder(root);
	printf("\n");
	preorder(root);
	printf("\n");
      }
    }

    return 0;
  }