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;
}