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>          // printf(), scanf()
#define MAX_N 500000

typedef struct node_tbl
{
	int key;
	int parent;
	int left;
	int right;
} node;

const int nil = -1;
node t[MAX_N + 1];
int root;
int ix;

int
find(int u, int k)
{
	while (u != nil && k != t[u].key)
	{
		if (k < t[u].key)
			u = t[u].left;
		else
			u = t[u].right;
	}

	return u;
}

void
insert(int k)
{
	int y = nil;
	int x = root;
	int z = ix++;

	t[z].key = k;
	t[z].left = nil;
	t[z].right = nil;

	while (x != nil)
	{
		y = x;
		if (t[z].key < t[x].key)
			x = t[x].left;
		else
			x = t[x].right;
	}

	t[z].parent = y;
	if (y == nil)
		root = z;
	else
	{
		if (t[z].key < t[y].key)
			t[y].left = z;
		else
			t[y].right = z;
	}
}

void
inorder(int u)
{
	if (u == nil)
		return;

	inorder(t[u].left);
	printf(" %d", t[u].key);
	inorder(t[u].right);
}

void
preorder(int u)
{
	if (u == nil)
		return;

	printf(" %d", t[u].key);
	preorder(t[u].left);
	preorder(t[u].right);
}

int
main(int argc, char** argv)
{
	char command[8];
	int n;
	int x;
	int i;

	root = nil;
	ix = 0;
	scanf("%d", &n);
	for (i = 0; i < n; ++i)
	{
		scanf("%s", command);
		switch (command[0])
		{
		case 'f':
			scanf("%d", &x);
			printf("%s\n", find(root, x) != nil ? "yes" : "no");
			break;

		case 'i':
			scanf("%d", &x);
			insert(x);
			break;

		case 'p':
		default:
			inorder(root);
			printf("\n");

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

	return 0;
}