📜  欧拉二叉树之旅

📅  最后修改于: 2021-06-26 15:38:58             🧑  作者: Mango

给定一个二叉树,每个节点最多可以有两个子节点,任务是找到二叉树的Euler环。欧拉巡回赛由指向树中最高节点的指针表示。如果树为空,则root的值为NULL。

例子:

方法:

欧拉树的巡回已经讨论过了,它可以应用于邻接表所表示的N元树。如果二进制树通过链接和节点以经典的结构化方式表示,则需要首先将其转换为邻接列表表示,然后,如果我们想应用原始文章中讨论的方法,则可以找到Euler巡回赛。但这增加了程序的空间复杂度。在此,本文讨论了一种广义的空间优化版本,该版本可以直接应用于以结构节点表示的二叉树。
此方法:
(1)无需使用Visited数组即可工作。
(2)恰好需要2 * N-1个顶点来存储Euler巡回。

C++
// C++ program to find euler tour of binary tree
#include 
using namespace std;
  
/* A tree node structure */
struct Node {
    int data;
    struct Node* left;
    struct Node* right;
};
  
/* Utility function to create a new Binary Tree node */
struct Node* newNode(int data)
{
    struct Node* temp = new struct Node;
    temp->data = data;
    temp->left = temp->right = NULL;
    return temp;
}
  
// Find Euler Tour
void eulerTree(struct Node* root, vector &Euler)
{
    // store current node's data
    Euler.push_back(root->data);
  
    // If left node exists
    if (root->left) 
    {
        // traverse left subtree
        eulerTree(root->left, Euler);
  
        // store parent node's data 
        Euler.push_back(root->data);
    }
  
    // If right node exists
    if (root->right) 
    {
        // traverse right subtree
        eulerTree(root->right, Euler); 
  
        // store parent node's data
        Euler.push_back(root->data);
    }
}
  
// Function to print Euler Tour of tree
void printEulerTour(Node *root)
{
    // Stores Euler Tour
    vector Euler; 
  
    eulerTree(root, Euler);
  
    for (int i = 0; i < Euler.size(); i++)
        cout << Euler[i] << " ";
}
  
/* Driver function to test above functions */
int main()
{
    // Constructing tree given in the above figure 
    Node* root = newNode(1);
    root->left = newNode(2);
    root->right = newNode(3);
    root->left->left = newNode(4);
    root->left->right = newNode(5);
    root->right->left = newNode(6);
    root->right->right = newNode(7);
    root->right->left->right = newNode(8);
  
    // print Euler Tour
    printEulerTour(root); 
  
    return 0;
}


Java
// Java program to find euler tour of binary tree
import java.util.*;
  
class GFG
{
  
    /* A tree node structure */
    static class Node
    {
        int data;
        Node left;
        Node right;
    };
  
    /* Utility function to create a new Binary Tree node */
    static Node newNode(int data)
    {
        Node temp = new Node();
        temp.data = data;
        temp.left = temp.right = null;
        return temp;
    }
  
    // Find Euler Tour
    static Vector eulerTree(Node root, Vector Euler)
    {
        // store current node's data
        Euler.add(root.data);
  
        // If left node exists
        if (root.left != null)
        {
            // traverse left subtree
            Euler = eulerTree(root.left, Euler);
  
            // store parent node's data
            Euler.add(root.data);
        }
  
        // If right node exists
        if (root.right != null)
        {
            // traverse right subtree
            Euler = eulerTree(root.right, Euler);
  
            // store parent node's data
            Euler.add(root.data);
        }
        return Euler;
    }
  
    // Function to print Euler Tour of tree
    static void printEulerTour(Node root) 
    {
        // Stores Euler Tour
        Vector Euler = new Vector();
  
        Euler = eulerTree(root, Euler);
  
        for (int i = 0; i < Euler.size(); i++)
            System.out.print(Euler.get(i) + " ");
    }
  
    /* Driver function to test above functions */
    public static void main(String[] args)
    {
        // Constructing tree given in the above figure
        Node root = newNode(1);
        root.left = newNode(2);
        root.right = newNode(3);
        root.left.left = newNode(4);
        root.left.right = newNode(5);
        root.right.left = newNode(6);
        root.right.right = newNode(7);
        root.right.left.right = newNode(8);
  
        // print Euler Tour
        printEulerTour(root);
  
    }
}
  
// This code is contributed by Rajput-Ji


C#
// C# program to find euler tour of binary tree
using System;
using System.Collections.Generic;
  
class GFG
{
  
    /* A tree node structure */
    public class Node
    {
        public int data;
        public Node left;
        public Node right;
    };
  
    /* Utility function to create a new Binary Tree node */
    static Node newNode(int data)
    {
        Node temp = new Node();
        temp.data = data;
        temp.left = temp.right = null;
        return temp;
    }
  
    // Find Euler Tour
    static List eulerTree(Node root, List Euler)
    {
        // store current node's data
        Euler.Add(root.data);
  
        // If left node exists
        if (root.left != null)
        {
            // traverse left subtree
            Euler = eulerTree(root.left, Euler);
  
            // store parent node's data
            Euler.Add(root.data);
        }
  
        // If right node exists
        if (root.right != null)
        {
            // traverse right subtree
            Euler = eulerTree(root.right, Euler);
  
            // store parent node's data
            Euler.Add(root.data);
        }
        return Euler;
    }
  
    // Function to print Euler Tour of tree
    static void printEulerTour(Node root) 
    {
        // Stores Euler Tour
        List Euler = new List();
  
        Euler = eulerTree(root, Euler);
  
        for (int i = 0; i < Euler.Count; i++)
            Console.Write(Euler[i] + " ");
    }
  
    /* Driver function to test above functions */
    public static void Main(String[] args)
    {
        // Constructing tree given in the above figure
        Node root = newNode(1);
        root.left = newNode(2);
        root.right = newNode(3);
        root.left.left = newNode(4);
        root.left.right = newNode(5);
        root.right.left = newNode(6);
        root.right.right = newNode(7);
        root.right.left.right = newNode(8);
  
        // print Euler Tour
        printEulerTour(root);
  
    }
}
  
// This code is contributed by 29AjayKumar


输出:
1 2 4 2 5 2 1 3 6 8 6 3 7 3 1

时间复杂度: O(2 * N-1),其中N是树中的节点数。
辅助空间: O(2 * N-1),其中N是树中的节点数。

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