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>

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

struct Node *root, *NIL;

void insert(int k){
    struct Node *y = NIL;
    struct Node *x = root;
    struct Node *z;

    z = (struct Node *)malloc(sizeof(struct Node));
    z->key = k;
    z->left = NIL;
    z->right = NIL;
    

    while(x!= NIL){
        y = x;
        if(z->key < x->key){
            x = x->left;
        } else{
            x = x->right;
        }
    }
    
    z->parent = y;
    if(y == NIL){
        root = z;
    } else{
        if( z->key < y->key){
            y->left = z;
        } else{
            y->right = z;
        }
    }

}

void find(int key){
    struct Node *x = root;

    while(x != NIL){
        if(x->key > key){
            x = x->left;
        } else if(x->key < key) {
            x = x->right;
        } else{
            printf("yes\n");
            return;
        }
    }

    printf("no\n");
}

void inorder(struct Node *u){
    if(u == NIL) return;
    inorder(u->left);
    printf(" %d", u->key);
    inorder(u->right);
}

void preorder(struct Node *u){
    if(u == NIL) return;
    printf(" %d", u->key);
    preorder(u->left);
    preorder(u->right);
}

int main(){
    int n, i, x;
    char com[10];
    scanf("%d", &n);

    for(i = 0; i < n; i++){
        scanf("%s", com);
        if(com[0] == 'i'){
            scanf("%d", &x);
            insert(x);
        } else if(com[0] == 'f'){
            scanf("%d", &x);
            find(x);
        } else{
            inorder(root);
            printf("\n");
            preorder(root);
            printf("\n");

        }
    }

    return 0;
}