• Binary Trees Aizu


    Binary Tree

    A rooted binary tree is a tree with a root node in which every node has at most two children.

    Your task is to write a program which reads a rooted binary tree T and prints the following information for each node u of T:

    node ID of u
    parent of u
    sibling of u
    the number of children of u
    depth of u
    height of u
    node type (root, internal node or leaf)
    If two nodes have the same parent, they are siblings. Here, if u and v have the same parent, we say u is a sibling of v (vice versa).

    The height of a node in a tree is the number of edges on the longest simple downward path from the node to a leaf.

    Here, the given binary tree consists of n nodes and evey node has a unique ID from 0 to n-1.

    Input

    The first line of the input includes an integer n, the number of nodes of the tree.

    In the next n lines, the information of each node is given in the following format:

    id left right

    id is the node ID, left is ID of the left child and right is ID of the right child. If the node does not have the left (right) child, the left(right) is indicated by -1.

    Output

    Print the information of each node in the following format:

    node id: parent = p , sibling = s , degree = deg, depth = dep, height = h, type

    p is ID of its parent. If the node does not have a parent, print -1.

    s is ID of its sibling. If the node does not have a sibling, print -1.

    deg, dep and h are the number of children, depth and height of the node respectively.

    type is a type of nodes represented by a string (root, internal node or leaf. If the root can be considered as a leaf or an internal node, print root.

    Please follow the format presented in a sample output below.

    Constraints

    1 ≤ n ≤ 25

    Sample Input 1

    9
    0 1 4
    1 2 3
    2 -1 -1
    3 -1 -1
    4 5 8
    5 6 7
    6 -1 -1
    7 -1 -1
    8 -1 -1

    Sample Output 1

    node 0: parent = -1, sibling = -1, degree = 2, depth = 0, height = 3, root
    node 1: parent = 0, sibling = 4, degree = 2, depth = 1, height = 1, internal node
    node 2: parent = 1, sibling = 3, degree = 0, depth = 2, height = 0, leaf
    node 3: parent = 1, sibling = 2, degree = 0, depth = 2, height = 0, leaf
    node 4: parent = 0, sibling = 1, degree = 2, depth = 1, height = 2, internal node
    node 5: parent = 4, sibling = 8, degree = 2, depth = 2, height = 1, internal node
    node 6: parent = 5, sibling = 7, degree = 0, depth = 3, height = 0, leaf
    node 7: parent = 5, sibling = 6, degree = 0, depth = 3, height = 0, leaf
    node 8: parent = 4, sibling = 5, degree = 0, depth = 2, height = 0, leaf
    在这里插入图片描述

    Reference

    Introduction to Algorithms, Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest, and Clifford Stein. The MIT Press.

    Code

    /*
                                    ^....0
                                   ^ .1 ^1^
                                   ..     01
                                  1.^     1.0
                                 ^ 1  ^    ^0.1
                                 1 ^        ^..^
                                 0.           ^ 0^
                                 .0            1 .^
                                 .1             ^0 .........001^
                                 .1               1. .111100....01^
                                 00                 11^        ^1. .1^
                                 1.^                              ^0  0^
                                   .^                                 ^0..1
                                   .1                                   1..^
                                 1 .0                                     ^  ^
                                  00.                                     ^^0.^
                                  ^ 0                                     ^^110.^
                              0   0 ^                                     ^^^10.01
                       ^^     10  1 1                                      ^^^1110.1
                       01     10  1.1                                      ^^^1111110
                       010    01  ^^                                        ^^^1111^1.^           ^^^
                       10  10^ 0^ 1                                            ^^111^^^0.1^       1....^
                        11     0                                               ^^11^^^ 0..  ....1^   ^ ^
                        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
                      0..^      ^0                              01               ^^^       ..      0..^
                     1 ..        .1                             ^.^              ^^^       1 ^  ^0001
                    ^  1.        00                              0.             ^^^        ^.0 ^.1
                    . 0^.        ^.^                             ^.^            ^^^         ..0.0
                   1 .^^.         .^                  1001        ^^            ^^^         . 1^
                   . ^ ^.         11                0.    1         ^           ^^          0.
                    0  ^.          0              ^0       1                   ^^^          0.
                  0.^  1.          0^             0       .1                   ^^^          ..
                  .1   1.          00            .        .1                  ^^^           ..
                 1      1.         ^.           0         .^                  ^^            ..
                 0.     1.          .^          .         0                                  .
                 .1     1.          01          .        .                                 ^ 0
                ^.^     00          ^0          1.       ^                                 1 1
                .0      00           .            ^^^^^^                                   .
                .^      00           01                                                    ..
               1.       00           10                                                   1 ^
              ^.1       00           ^.                                            ^^^    .1
              ..        00            .1                                        1..01    ..
             1.1         00           1.                                       ..^      10
            ^ 1^         00           ^.1                                      0 1      1
            .1           00            00                                       ^  1   ^
             .           00            ^.^                                        10^  ^^
           1.1           00             00                                              10^
           ..^           1.             ^.                                               1.
          0 1            ^.              00                 00                            .^
            ^            ^.              ^ 1                00   ^0000^     ^               01
         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
       ..                                                          1.1            ^0.0
      ^ 0                                                           1..01^^100000..0^
      1 1                                                            ^ 1 ^^1111^ ^^
      0 ^                                                             ^ 1      1000^
      .1                                                               ^.^     .   00
      ..                                                                1.1    0.   0
      1.                                                                  .    1.   .^
      1.                                                                 1    1.   ^0
     ^ .                                                                 ^.1 00    01
     ^.0                                                                  001.     .^
     */
    // Virtual_Judge —— Binary Trees Aizu - ALDS1_7_B.cpp created by VB_KoKing on 2019-05-08:09.
    /* 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>
    
    #define MAX 10007
    #define NIL -1
    
    using namespace std;
    
    struct Node {
        int parent, left, right;
    };
    
    Node T[MAX];
    int n, D[MAX], H[MAX];
    
    void set_depth(int u, int d) {
        if (u == NIL) return;
        D[u] = d;
        set_depth(T[u].left, d + 1);
        set_depth(T[u].right, d + 1);
    }
    
    int set_height(int u) {
        int h1 = 0, h2 = 0;
        if (T[u].left != NIL)
            h1 = set_height(T[u].left) + 1;
        if (T[u].right != NIL)
            h2 = set_height(T[u].right) + 1;
        return H[u] = max(h1, h2);
    }
    
    int get_sibling(int u) {
        if (T[u].parent == NIL) return NIL;
        if (T[T[u].parent].left != u && T[T[u].parent].left != NIL)
            return T[T[u].parent].left;
        if (T[T[u].parent].right != u && T[T[u].parent].right != NIL)
            return T[T[u].parent].right;
        return NIL;
    }
    
    void print(int u) {
        cout << "node " << u << ": ";
        cout << "parent = " << T[u].parent << ", ";
        cout << "sibling = " << get_sibling(u) << ", ";
    
        int deg = 0;
        if (T[u].left != NIL) deg++;
        if (T[u].right != NIL) deg++;
        cout << "degree = " << deg << ", ";
    
        cout << "depth = " << D[u] << ", ";
        cout << "height = " << H[u] << ", ";
    
        if (T[u].parent == NIL)
            cout << "root" << endl;
        else if (T[u].left == NIL && T[u].right == NIL)
            cout << "leaf" << endl;
        else
            cout << "internal node" << endl;
    
    }
    
    int main() {
        int v, l, r, root = 0;
        cin >> n;
    
        for (int i = 0; i < n; i++) {
            T[i].parent = NIL;
        }
    
        for (int i = 0; i < n; i++) {
            cin >> v >> l >> r;
            T[v].left = l;
            T[v].right = r;
            if (l != NIL) T[l].parent = v;
            if (r != NIL) T[r].parent = v;
        }
    
        for (int i = 0; i < n; i++)
            if (T[i].parent == NIL)
                root = i;
    
        set_depth(root, 0);
        set_height(root);
    
        for (int i = 0; i < n; i++)
            print(i);
    
        return 0;
    }
    
  • 相关阅读:
    七牛云上传图片
    找到当前字符串中最后一个/并获取之后的字符串
    jquery正则表达式验证:验证身份证号码
    apply()与call()的区别
    js 判断字符串是否包含某字符串,String对象中查找子字符,indexOf
    改变父元素的透明度,不影响子元素的透明度—css
    c实现生产者消费者问题。 windows下。
    python基础练习 dict切片
    html+css test1
    codewars[7]-python Friend or Foe?
  • 原文地址:https://www.cnblogs.com/AlexKing007/p/12338337.html
Copyright © 2020-2023  润新知