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_ Node;

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

static Node* root;

static void insert(int key)
{
    Node* node = (Node*)malloc(sizeof(Node));
    node->key = key;
    node->parent = NULL;
    node->left = NULL;
    node->right = NULL;
    
    Node* parent = NULL;
    Node* cur = root;
    while (cur != NULL) {
        parent = cur;
        if (node->key < cur->key) {
            cur = cur->left;
        } else {
            cur = cur->right;
        }
    }
    
    node->parent = parent;
    
    if (parent == NULL) {
        root = node;
    } else {
        if (node->key < parent->key) {
            parent->left = node;
        } else {
            parent->right = node;
        }
    }
}

static Node* find(int key)
{
    Node* node = root;
    while (node != NULL && node->key != key) {
        if (key < node->key) {
            node = node->left;
        } else {
            node = node->right;
        }
    }
    return node;
}

static void printInorder(Node* node)
{
    if (node == NULL)
        return;
    printInorder(node->left);
    printf(" %d", node->key);
    printInorder(node->right);
}

static void printPreorder(Node* node)
{
    if (node == NULL)
        return;
    printf(" %d", node->key);
    printPreorder(node->left);
    printPreorder(node->right);
}

static void printTree()
{
    printInorder(root);
    printf("\n");
    printPreorder(root);
    printf("\n");
}

static void freeTree(Node* node)
{
    if (node->left != NULL) {
        freeTree(node->left);
    }
    if (node->right != NULL) {
        freeTree(node->right);
    }
    free(node);
}

int main()
{
    int n;
    
    scanf("%d", &n);
    for (int i = 0; i < n; i++) {
        char command[10];
        scanf("%s", command);
        
        if (command[0] == 'i') {
            int key;
            scanf("%d", &key);
            insert(key);
        } else if (command[0] == 'f') {
            int key;
            scanf("%d", &key);
            Node* node = find(key);
            printf((node != NULL) ? "yes\n" : "no\n");
        } else if (command[0] == 'p') {
            printTree();
        }
    }
    
    freeTree(root);
    return 0;
}