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>

#define N 500000

int n;

struct Tree{
	int key;
	struct Tree *parent,*left,*right;
};
struct Tree *NILL,*root;

void find(struct Tree *root,int key){
	int flag =0;
	while(root!=NILL){
		if(root->key > key)root = root->left;
		else if(root->key == key) {
			printf("yes\n");
			flag =1;
			break;
		}
		else root = root->right;
	}
	if(!flag)printf("no\n");
}

void insert(struct Tree *z){
	struct Tree *x,*y;
	y = NILL;
	x = root;
	while(x != NILL){
		y = x;
		if(z->key < x->key) x = x->left;
		else x = x->right;
	}
	z->parent = y;
	if(y == NILL)  {
		root = z;
	}
	else if(z->key < y->key) y->left = z;
	else y->right = z;
}

void mid(struct Tree *root){
	if(root == NILL) return;
	mid(root->left);
	printf(" %d", root->key);
	mid(root->right);
}

void inorder(struct Tree *root){
	if(root == NILL) return;
	printf(" %d",root->key);
	inorder(root->left);
	inorder(root->right);
}

void print(){
	struct Tree *r;
	r = root;
	mid(r);
	printf("\n");
	inorder(r);
	printf("\n");
}

int main(){
	struct Tree *t;
	scanf("%d",&n);
	int i;
	int key;
	char instruct[7];
	for(i=0;i<n;i++){
		scanf("%s",instruct);
		if(!strcmp(instruct,"insert")){
			scanf("%d",&key);
			t = (struct Tree *)malloc(sizeof(struct Tree));
			t->key = key;
			t->right = NILL;
			t->left = NILL;
			//T[i]->key  = key;
			insert(t);
		}
		else if(!strcmp(instruct,"find")){
			scanf("%d",&key);
			find(root,key);
		}
		else print();
	}
	return 0;
}