Problem:
MathJax.Hub.Config({ tex2jax: { inlineMath: [["$","$"], ["\\(","\\)"]], processEscapes: true }});

Binary Search Tree I

Search trees are data structures that support dynamic set operations including insert, search, delete and so on. Thus a search tree can be used both as a dictionary and as a priority queue.

Binary search tree is one of fundamental search trees. The keys in a binary search tree are always stored in such a way as to satisfy the following binary search tree property:

Let $x$ be a node in a binary search tree. If $y$ is a node in the left subtree of $x$, then $y.key \leq x.key$. If $y$ is a node in the right subtree of $x$, then $x.key \leq y.key$.

The following figure shows an example of the binary search tree.

For example, keys of nodes which belong to the left sub-tree of the node containing 80 are less than or equal to 80, and keys of nodes which belong to the right sub-tree are more than or equal to 80. The binary search tree property allows us to print out all the keys in the tree in sorted order by an inorder tree walk.

A binary search tree should be implemented in such a way that the binary search tree property continues to hold after modifications by insertions and deletions. A binary search tree can be represented by a linked data structure in which each node is an object. In addition to a key field and satellite data, each node contains fields left, right, and p that point to the nodes corresponding to its left child, its right child, and its parent, respectively.

To insert a new value $v$ into a binary search tree $T$, we can use the procedure insert as shown in the following pseudo code. The insert procedure is passed a node $z$ for which $z.key = v$, $z.left = NIL$, and $z.right = NIL$. The procedure modifies $T$ and some of the fields of $z$ in such a way that $z$ is inserted into an appropriate position in the tree.

1 insert(T, z)
2 y = NIL // parent of x
3 x = 'the root of T'
4 while x ≠ NIL
5 y = x // set the parent
6 if z.key < x.key
7 x = x.left // move to the left child
8 else
9 x = x.right // move to the right child
10 z.p = y
11
12 if y == NIL // T is empty
13 'the root of T' = z
14 else if z.key < y.key
15 y.left = z // z is the left child of y
16 else
17 y.right = z // z is the right child of y

Write a program which performs the following operations to a binary search tree $T$.

insert $k$: Insert a node containing $k$ as key into $T$.

print: Print the keys of the binary search tree by inorder tree walk and preorder tree walk respectively.

You should use the above pseudo code to implement the insert operation. $T$ is empty at the initial state.

Input

In the first line, the number of operations $m$ is given. In the following $m$ lines, operations represented by insert $k$ or print are given.

Output

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

8
insert 30
insert 88
insert 12
insert 1
insert 20
insert 17
insert 25
print

Sample Output 1

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>
#include <string.h>

// 構造体を宣言
struct Node {
    // 自分のなまえ
    int key;
    // 子(右) 子(左) 親 のポインタ
    struct Node *right, *left, *parent;
};
typedef struct Node Node;

Node *root, *NIL;

void insert(int k) {
    // 構造体yに構造体NILを代入
    Node *y = NIL;
    // 構造体xに構造体rootを代入
    Node *x = root;
    // 構造体zを出す
    Node *z;

    // 構造体zと構造体分のメモリを確保して新しく宣言した構造体を対応させる
    z = (Node *)malloc(sizeof(Node));
    // 構造体zのkeyにkを入力する
    z->key   = k;
    // 構造体zの子を両方NILにする
    z->left  = NIL;
    z->right = NIL;

    // 一番下に行くまで(番兵が出るまで)分岐して行く
    while(x != NIL) {
        // 構造体yに構造体xを代入する
        y = x;
        // 参照するキーより小さかったら左に、大きかったら右の子へ移動する
        if(z->key < x->key) {
            x = x->left;
        } else {
            x = x->right;
        }
    }

    // 構造体yを構造体zのparentに入れる
    // z->parent = y;
    // 構造体yがNILだった場合(木が空の場合)rootにzを代入する
    if(y == NIL) {
        root = z;
    }
    // 挿入するところのノードと挿入するやつでどっちが大きいかによって左右分ける
    // 挿入するやつの方が小さかったら左、大きかったら右
    else {
        if( z->key < y->key) {
            y->left = z;
        } else {
            y->right = z;
        }
    }
}

// 中間順巡回 (Inorder Tree Walk)
void inorder(Node *u) {
    if(u == NIL) return;
    inorder(u->left);
    printf(" %d",u->key);
    inorder(u->right);
}
// 先行順巡回 (Preorder Tree Walk)
void preorder(Node *u) {
    if(u == NIL) return;
    printf(" %d",u->key);
    preorder(u->left);
    preorder(u->right);
}

int main(void) {
    int n, i, x;
    // 文字列を取得するための配列を宣言
    char com[16];
    // ノードの個数を取得
    scanf("%d",&n);

    for(i=0;i<n;i++) {
        // 文字列を取得
        scanf("%s",com);
        // 文字列がinsertだった場合
        if(!strcmp(com,"insert")) {
            // 挿入する数字を取得
            scanf("%d",&x);
            // 挿入する
            insert(x);
        }
        // 文字列がprintだった場合
        else if(!strcmp(com,"print")) {
            // 中間順巡回(inorder tree walk)を出力
            inorder(root);
            printf("\n");
            // 先行順巡回(preorder tree walk)を出力
            preorder(root);
            printf("\n");
        }
    }

    return 0;
}

/*
8
insert 30
insert 88
insert 12
insert 1
insert 20
insert 17
insert 25
print
*/