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<iostream>
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i,n) for(int i=0;i<n;i++)

/* 大文字を小文字に変換 */
char tolower(char c) {return (c + 0x20);}
/* 小文字を大文字に変換 */
char toupr(char c) {return (c - 0x20);}

// if('A'<=s[i] && s[i]<='Z') s[i] += 'a'-'A';

/*
string s = "abcdefg"
s.substr(4) "efg"
s.substr(0,3) "abc"
s.substr(2,4) "cdef"
*/

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

Node *root, *NIL;

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

    // z = malloc(sizeof(struct node));  // node構造体サイズのメモリ領域をとる
    z = (Node *)malloc(sizeof(Node));
    z->key = k;   // ポインタ変数にアクセスする時はアロー演算子
    z->left = NIL;
    z->right = NIL;
    
    while(x != NIL){
        y = x;    // 最終的なzの親を決定する
        if(z->key < x->key) x = x->left;
        else x = x->right;
    }
    z->parent = y;
    if(y == NIL){
        root = z; // 親要素が空であれば、zが根となる
    }else{
        if(z->key < y->key){
            y->left = z;
        }else{
            y->right = z;
        }
    }
}

bool find(int k){
    Node *x = root;
    Node *y = NIL;
    while(x != NIL){
        if(x->key == k){
            return true;     
        }else if(x->key < k){
            x = x->right;
        }else{
            x = x->left;
        }
    }
    return false;
}

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

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

int main()
{
    int n, i, x;
    string com;
    scanf("%d", &n);

    rep(i, n){
        cin >> com;
        if(com == "insert"){
            scanf("%d", &x);
            insert(x);
        }else if(com == "print"){
            inorder(root);
            printf("\n");
            preorder(root);
            printf("\n");
        }else if(com == "find"){
            scanf("%d", &x);
            if(find(x)) cout << "yes" << endl;
            else cout << "no" << endl;
        }
    }
}