二叉树（数组实现）

给定一个表示树的数组，数组索引是树节点中的值，数组值给出该特定索引（或节点）的父节点。根节点索引的值将始终为 -1，因为根没有父节点。从这个给定的表示构造给定二叉树的标准链接表示。请参考以了解如何从给定的父数组表示构造二叉树。

表示方式：

树可以用以下两种方式表示：

1. 动态节点表示（链接表示）。
2. 数组表示（顺序表示）。

现在，我们将讨论树的顺序表示。为了使用数组来表示树，节点的编号可以从 0-(n-1) 或 1-n 开始，请考虑下图：

插图：

A(0)    
     /   \
    B(1)  C(2)  
  /   \      \
 D(3)  E(4)   F(6) 
OR,
      A(1)    
     /   \
    B(2)  C(3)  
  /   \      \
 D(4)  E(5)   F(7)  

程序：

Note: father, left_son and right_son are the values of indices of the array.

案例一： （0—n-1）

if (say)father=p; 
then left_son=(2*p)+1; 
and right_son=(2*p)+2;

案例 2： 1—n

if (say)father=p; 
then left_son=(2*p); 
and right_son=(2*p)+1; 

执行：

例子

C++

// C++ implementation of tree using array
// numbering starting from 0 to n-1.
#include
using namespace std;
char tree[10];
int root(char key) {
  if (tree[0] != '\0')
    cout << "Tree already had root";
  else
    tree[0] = key;
  return 0;
}
 
int set_left(char key, int parent) {
  if (tree[parent] == '\0')
    cout << "\nCan't set child at"
    << (parent * 2) + 1
    << " , no parent found";
  else
    tree[(parent * 2) + 1] = key;
  return 0;
}
 
int set_right(char key, int parent) {
  if (tree[parent] == '\0')
    cout << "\nCan't set child at"
    << (parent * 2) + 2
    << " , no parent found";
  else
    tree[(parent * 2) + 2] = key;
  return 0;
}
 
int print_tree() {
  cout << "\n";
  for (int i = 0; i < 10; i++) {
    if (tree[i] != '\0')
      cout << tree[i];
    else
      cout << "-";
  }
  return 0;
}
 
// Driver Code
int main() {
  root('A');
  //insert_left('B',0);
  set_right('C', 0);
  set_left('D', 1);
  set_right('E', 1);
  set_right('F', 2);
  print_tree();
  return 0;
}

Java

// JAVA implementation of tree using array
// numbering starting from 0 to n-1.
 
// Importing required classes
import java.io.*;
import java.lang.*;
import java.util.*;
 
// Class 1
// Helper class (Node class)
class Tree {
 
    // Main driver method
    public static void main(String[] args)
    {
 
        // Creating object of class 2 inside main() method
        Array_imp obj = new Array_imp();
 
        // Setting root node
        obj.Root("A");
 
        // obj.set_Left("B", 0);
        obj.set_Right("C", 0);
        obj.set_Left("D", 1);
        obj.set_Right("E", 1);
        obj.set_Left("F", 2);
        obj.print_Tree();
    }
}
 
// Class 2
// Helper class
class Array_imp {
 
    // Member variables of this class
    static int root = 0;
    static String[] str = new String[10];
 
    // Method 1
    // Creating root node
    public void Root(String key) { str[0] = key; }
 
    // Method 2
    // Creating left son of root
    public void set_Left(String key, int root)
    {
        int t = (root * 2) + 1;
 
        if (str[root] == null) {
            System.out.printf(
                "Can't set child at %d, no parent found\n",
                t);
        }
        else {
            str[t] = key;
        }
    }
 
    // Method 3
    // Creating right son of root
    public void set_Right(String key, int root)
    {
        int t = (root * 2) + 2;
 
        if (str[root] == null) {
            System.out.printf(
                "Can't set child at %d, no parent found\n",
                t);
        }
        else {
            str[t] = key;
        }
    }
 
    // Method 4
    // To print our tree
    public void print_Tree()
    {
 
        // Iterating using for loop
        for (int i = 0; i < 10; i++) {
            if (str[i] != null)
                System.out.print(str[i]);
            else
                System.out.print("-");
        }
    }
}

C#

// C# implementation of tree using array
// numbering starting from 0 to n-1.
using System;
 
public class Tree {
    public static void Main(String[] args)
    {
        Array_imp obj = new Array_imp();
        obj.Root("A");
        // obj.set_Left("B", 0);
        obj.set_Right("C", 0);
        obj.set_Left("D", 1);
        obj.set_Right("E", 1);
        obj.set_Left("F", 2);
        obj.print_Tree();
    }
}
 
class Array_imp {
    static int root = 0;
    static String[] str = new String[10];
 
    /*create root*/
    public void Root(String key)
    {
        str[0] = key;
    }
 
    /*create left son of root*/
    public void set_Left(String key, int root)
    {
        int t = (root * 2) + 1;
 
        if (str[root] == null) {
            Console.Write("Can't set child at {0}, no parent found\n", t);
        }
        else {
            str[t] = key;
        }
    }
 
    /*create right son of root*/
    public void set_Right(String key, int root)
    {
        int t = (root * 2) + 2;
 
        if (str[root] == null) {
            Console.Write("Can't set child at {0}, no parent found\n", t);
        }
        else {
            str[t] = key;
        }
    }
 
    public void print_Tree()
    {
        for (int i = 0; i < 10; i++) {
            if (str[i] != null)
                Console.Write(str[i]);
            else
                Console.Write("-");
        }
    }
}
 
// This code contributed by Rajput-Ji

Python3

# Python3 implementation of tree using array
# numbering starting from 0 to n-1.
tree = [None] * 10
 
 
def root(key):
    if tree[0] != None:
        print("Tree already had root")
    else:
        tree[0] = key
 
 
def set_left(key, parent):
    if tree[parent] == None:
        print("Can't set child at", (parent * 2) + 1, ", no parent found")
    else:
        tree[(parent * 2) + 1] = key
 
 
def set_right(key, parent):
    if tree[parent] == None:
        print("Can't set child at", (parent * 2) + 2, ", no parent found")
    else:
        tree[(parent * 2) + 2] = key
 
 
def print_tree():
    for i in range(10):
        if tree[i] != None:
            print(tree[i], end="")
        else:
            print("-", end="")
    print()
 
 
# Driver Code
root('A')
set_right('C', 0)
set_left('D', 1)
set_right('E', 1)
set_right('F', 2)
print_tree()
 
# This code is contributed by Gaurav Kumar Tailor

输出

Can't set child at3 , no parent found
Can't set child at4 , no parent found
A-C---F---