Binary Search Tree II Aizu - ALDS1_8_B

Write a program which performs the following operations to a binary search tree TT by adding the find operation to A: Binary Search Tree I.

insert kk: Insert a node containing kk as key into TT.
find kk: Report whether TT has a node containing kk.
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 mm is given. In the following mm lines, operations represented by insert kk, find kk or print are given.

Output

For each find kk operation, print “yes” if TT has a node containing kk, “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 500,000\leq 500,000
The number of print operations 10\leq 10.
2,000,000,000key2,000,000,000-2,000,000,000 \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.

Code

/*
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                   ^^     10  1 1                                      ^^^1110.1
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                   10  10^ 0^ 1                                            ^^111^^^0.1^       1....^
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                    1.     0^                                               ^11^^^ ^ 1 111^     ^ 0.
                   10   00 11                                               ^^^^^   1 0           1.
                   0^  ^0  ^0                                                ^^^^    0            0.
                   0^  1.0  .^                                               ^^^^    1 1          .0
                   ^.^  ^^  0^                             ^1                ^^^^     0.         ^.1
                   1 ^      11                             1.                ^^^     ^ ^        ..^
                  ^..^      ^1                             ^.^               ^^^       .0       ^.0
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                . 0^.        ^.^                             ^.^            ^^^         ..0.0
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                0  ^.          0              ^0       1                   ^^^          0.
              0.^  1.          0^             0       .1                   ^^^          ..
              .1   1.          00            .        .1                  ^^^           ..
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             0.     1.          .^          .         0                                  .
             .1     1.          01          .        .                                 ^ 0
            ^.^     00          ^0          1.       ^                                 1 1
            .0      00           .            ^^^^^^                                   .
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           1.       00           10                                                   1 ^
          ^.1       00           ^.                                            ^^^    .1
          ..        00            .1                                        1..01    ..
         1.1         00           1.                                       ..^      10
        ^ 1^         00           ^.1                                      0 1      1
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      0 1            ^.              00                 00                            .^
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     1 0             ^.               00.0^              ^00000   1.00.1              11
     . 1              0               1^^0.01                      ^^^                01
      .^              ^                1   1^^                                       ^.^
    1 1                                                                              0.
    ..                                                                              1 ^
     1                                                                               1
   ^ ^                                                                             .0
   1                                                                             ^ 1
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  ^ 0                                                           1..01^^100000..0^
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  0 ^                                                             ^ 1      1000^
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 */
// Virtual_Judge —— Binary Search Tree I Aizu - ALDS1_8_A.cpp created by VB_KoKing on 2019-05-09:18.
/* Procedural objectives:

 Variables required by the program:

 Procedural thinking:

 Functions required by the program:

 Determination algorithm:

 Determining data structure:


*/
/* My dear Max said:
"I like you,
So the first bunch of sunshine I saw in the morning is you,
The first gentle breeze that passed through my ear is you,
The first star I see is also you.
The world I see is all your shadow."

FIGHTING FOR OUR FUTURE!!!
*/
#include <iostream>
#include <cstdlib>

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

Node *root, *NIL;

Node *find(Node *u,int k)
{
    while (u!=NIL&&k!=u->key)
    {
        if (k<u->key) u=u->left;
        else u=u->right;
    }
    return u;
}

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

    z = (Node *) malloc(sizeof(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;
    }
}

//前序遍历
void pre_parse(Node *u) {
    if (u == NIL) return;
    cout << " " << u->key;
    pre_parse(u->left);
    pre_parse(u->right);
}

//中序遍历
void in_parse(Node *u) {
    if (u == NIL) return;
    in_parse(u->left);
    cout << " " << u->key;
    in_parse(u->right);
}

int main() {
    int n;
    string com;
    cin >> n;
    for (int i = 0; i < n; i++) {
        cin >> com;
        if (com == "insert") {
            int x;
            cin >> x;
            insert(x);
        } else if (com == "print") {
            in_parse(root);
            cout << endl;
            pre_parse(root);
            cout << endl;
        } else if (com=="find"){
            int x;
            cin>>x;
            Node *t=find(root,x);
            if (t!=NIL) cout<<"yes"<<endl;
            else cout<<"no"<<endl;
        }
    }
    return 0;
}

posted @ 2019-05-09 20:12  AlexKing007  阅读(96)  评论(0编辑  收藏  举报