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 *r,*l,*p;
};
struct node *NIL,*root;
int a=0,b=0;
void insert(int k) {
	struct node *x=root,*y=NIL,*z;
	z=(struct node*)malloc(sizeof(struct node));
	z->key=k;
	z->l=NIL;
	z->r=NIL;
	while(x!=NIL) {
		y=x;
		if(k<x->key)
			x=x->l;
		else
			x=x->r;
	}
	z->p=y;
	if(y==NIL) {
		root=z;
	} else if(z->key<y->key)
		y->l=z;
	else
		y->r=z;
}
void inOrder(struct node *p) {
	if(p==NIL)
	return;
	inOrder(p->l);
    printf(" %d",p->key);
	inOrder(p->r);
}
void preOrder(struct node *p) {
	if(p==NIL)
	return;
	printf(" %d",p->key);
	preOrder(p->l);
	preOrder(p->r);
}
int print(struct node *p) {
	a=0;
	b=0;
	inOrder(p);
	printf("\n");
	preOrder(p);
	printf("\n");
	return a;
}
struct node *find(struct node *x,int k){
	while(x!=NIL&&x->key!=k){
		if(k>x->key)
		x=x->r;
		else
		x=x->l;
	}
	return x;
}
int main() {
	int m,i,j,o=0,k;
	char str[10];
	int len[11];
	struct node *x;
	scanf("%d",&m);
	for(i=0; i<m; i++) {
		scanf("%s",str);
		if(str[0]=='i') {
			scanf("%d",&k);
			insert(k);
		} else if(str[0]=='p'){
			print(root);
		}
		else{
			scanf("%d",&k);
			x=find(root,k);
			if(x!=NIL)
			printf("yes\n");
			else
			printf("no\n");
		}
	}//输入
	return 0;
}