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>

typedef struct node_t {
	int key;
	struct node_t *parent;
	struct node_t *left;
	struct node_t *right;
} node;

node* newNode(int);
void print(node *);
void inorder(node *);
void preorder(node *);
void insert(int, node *);
node* find(int, node *);
void free_tree(node *);

int main(void)
{
	int i, m, k;
	char com[10];
	node *tree_root;
	
	tree_root = newNode(0);
	
	scanf("%d",&m);
	for (i=0; i<m; i++) {
		scanf("%s",com);
		if(com[0] == 'p') {
			print(tree_root->left);
		} else {
			scanf("%d",&k);
			if(com[0] == 'i') {
				insert(k, tree_root);
			} else if(com[0] == 'f') {
				if(find(k, tree_root) != NULL) {
					printf("yes\n");
				} else {
					printf("no\n");
				}
			}
		}
	}
	
	free_tree(tree_root);
	
	return 0;
}

node* newNode(int key)
{
	node *tmp;
	
	tmp = (node *)malloc(sizeof(node));
	if (tmp == NULL) exit(0);
	
	tmp->key = key;
	tmp->parent = NULL;
	tmp->left = NULL;
	tmp->right = NULL;
	
	return tmp;
}

void print(node *root)
{
	inorder(root);
	printf("\n");
	preorder(root);
	printf("\n");
}

void inorder(node *node)
{
	
	if(node->left != NULL) inorder(node->left);
	printf(" %d",node->key);
	if(node->right != NULL) inorder(node->right);
}

void preorder(node *node)
{
	printf(" %d",node->key);
	if(node->left != NULL) preorder(node->left);
	if(node->right != NULL) preorder(node->right);
}

void insert(int k, node *tree_root)
{
	node *x, *y, *z;
	
	z = newNode(k);
	y = tree_root;
	x = tree_root->left;
	while (x != NULL) {
		y = x;
		if(z->key < x->key) {
			x = x->left;
		} else {
			x = x->right;
		}
	}
	z->parent = y;
	if(y == tree_root || z->key < y->key) {
		y->left = z;
	} else {
		y->right = z;
	}
}

node* find(int k, node *tree_root) 
{
	node *x;
	
	x = tree_root->left;
	while (x->key != k) {
		if(x->key > k) {
			if(x->left == NULL) return NULL;
			x = x->left;
		} else {
			if(x->right == NULL) return NULL;
			x = x->right;
		}
	}
	return x;
}

void free_tree(node *root)
{
	if(root->left != NULL) free_tree(root->left);
	if(root->right != NULL) free_tree(root->right);
	free(root);
}