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>
#include <stdlib.h>


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

struct Node *root , *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(void)
{
	int n , i , x;
	char com[100];
	char str1[] = "insert";
	char str2[] = "print";
	scanf("%d",&n);
	
	for( i = 0; i < n; i++)
	{
		while(scanf("%s",com) != EOF)
		
		if( com[0] == 'f')
		{
			scanf("%d",&x);
			struct Node *t = find(root,x);
			if(t != NIL) printf("yes\n");
			else printf("no\n");
		}
		else if(strcmp(com,str1)==0)
		{
			scanf("%d",&x);
			insert(x);
		}
		else if(strcmp(com,str2)==0)
		{
			inorder(root);
			printf("\n");
			preorder(root);
			printf("\n");
		}
	}
	return 0;
}