检查二叉树的节点是否形成算术、几何或调和级数

给定一棵二叉树，任务是检查这棵树中的节点是否形成等差级数、等比级数或调和级数。
例子：

Input:               
                      4
                    /   \
                   2     16
                  / \    / \
                 1   8  64  32
Output: Geometric Progression
Explanation:
The nodes of the binary tree can be used
to form a Geometric Progression as follows - 
{1, 2, 4, 8, 16, 32, 64}

Input:          
                 15
               /   \
              5     10
             / \      \
            25   35   20
Output: Arithmetic Progression
Explanation:
The nodes of the binary tree can be used
to form a Arithmetic Progression as follows - 
{5, 10, 15, 20, 25, 35}

方法：思想是使用Level-order Traversal遍历二叉树并将所有节点存储在一个数组中，然后检查该数组是否可以用于形成算术、几何或调和级数。

• 要检查序列是否在等差数列中，请对序列进行排序并检查数组中连续元素之间的公共差异是否相同。
• 要检查一个序列是否呈几何级数，对序列进行排序并检查数组中连续元素之间的公比是否相同。
• 要检查一个序列是否在调和级数中，找到每个元素的倒数，然后对数组进行排序并检查连续元素之间的公共差是否相同。

下面是上述方法的实现：

C++

// C++ implementation to check that
// nodes of binary tree form AP/GP/HP
 
#include 
using namespace std;
 
// Structure of the
// node of the binary tree
struct Node {
    int data;
    struct Node *left, *right;
};
 
// Function to find the size
// of the Binary Tree
int size(Node* node)
{
    // Base Case
    if (node == NULL)
        return 0;
    else
        return (size(node->left) + 1 + size(node->right));
}
 
// Function to check if the permutation
// of the sequence form Arithmetic Progression
bool checkIsAP(double arr[], int n)
{
    // If the sequence contains
    // only one element
    if (n == 1)
        return true;
 
    // Sorting the array
    sort(arr, arr + n);
 
    double d = arr[1] - arr[0];
 
    // Loop to check if the sequence
    // have same common difference
    // between its consecutive elements
    for (int i = 2; i < n; i++)
        if (arr[i] - arr[i - 1] != d)
            return false;
 
    return true;
}
 
// Function to check if the permutation
// of the sequence form
// Geometric progression
bool checkIsGP(double arr[], int n)
{
    // Condition when the length
    // of the sequence is 1
    if (n == 1)
        return true;
 
    // Sorting the array
    sort(arr, arr + n);
    double r = arr[1] / arr[0];
 
    // Loop to check if the the
    // sequence have same common
    // ratio in consecutive elements
    for (int i = 2; i < n; i++) {
        if (arr[i] / arr[i - 1] != r)
            return false;
    }
    return true;
}
 
// Function to check if the permutation
// of the sequence form
// Harmonic Progression
bool checkIsHP(double arr[], int n)
{
    // Condition when length of
    // sequence in 1
    if (n == 1) {
        return true;
    }
    double rec[n];
 
    // Loop to find the reciprocal
    // of the sequence
    for (int i = 0; i < n; i++) {
        rec[i] = ((1 / arr[i]));
    }
 
    // Sorting the array
    sort(rec, rec + n);
    double d = (rec[1]) - (rec[0]);
 
    // Loop to check if the common
    // difference of the sequence is same
    for (int i = 2; i < n; i++) {
        if (rec[i] - rec[i - 1] != d) {
            return false;
        }
    }
    return true;
}
 
// Function to check if the nodes
// of the Binary tree forms AP/GP/HP
void checktype(Node* root)
{
 
    int n = size(root);
    double arr[n];
    int i = 0;
 
    // Base Case
    if (root == NULL)
        return;
 
    // Create an empty queue
    // for level order traversal
    queue q;
 
    // Enqueue Root and initialize height
    q.push(root);
 
    // Loop to traverse the tree using
    // Level order Traversal
    while (q.empty() == false) {
        Node* node = q.front();
        arr[i] = node->data;
        i++;
        q.pop();
 
        // Enqueue left child
        if (node->left != NULL)
            q.push(node->left);
 
        // Enqueue right child
        if (node->right != NULL)
            q.push(node->right);
    }
 
    int flag = 0;
 
    // Condition to check if the
    // sequence form Arithmetic Progression
    if (checkIsAP(arr, n)) {
        cout << "Arithmetic Progression"
             << endl;
        flag = 1;
    }
 
    // Condition to check if the
    // sequence form Geometric Progression
    else if (checkIsGP(arr, n)) {
        cout << "Geometric Progression"
             << endl;
        flag = 1;
    }
 
    // Condition to check if the
    // sequence form Geometric Progression
    else if (checkIsHP(arr, n)) {
        cout << "Geometric Progression"
             << endl;
        flag = 1;
    }
    else if (flag == 0) {
        cout << "No";
    }
}
 
// Function to create new node
struct Node* newNode(int data)
{
    struct Node* node = new Node;
    node->data = data;
    node->left = node->right = NULL;
    return (node);
}
 
// Driver Code
int main()
{
    /* Constructed Binary tree is:
             1
            / \
           2   3
          / \   \
         4   5   8
                / \
                6  7
    */
    struct Node* root = newNode(1);
    root->left = newNode(2);
    root->right = newNode(3);
    root->left->left = newNode(4);
    root->left->right = newNode(5);
    root->right->right = newNode(8);
    root->right->right->left = newNode(6);
    root->right->right->right = newNode(7);
 
    checktype(root);
 
    return 0;
}

Java

// Java implementation to check that
// nodes of binary tree form AP/GP/HP
import java.util.*;
 
class GFG {
    // Structure of the
    // node of the binary tree
    static class Node {
        int data;
        Node left, right;
 
        Node(int data) {
            this.data = data;
            this.left = this.right = null;
        }
    };
 
    // Function to find the size
    // of the Binary Tree
    static int size(Node node) {
        // Base Case
        if (node == null)
            return 0;
        else
            return (size(node.left) + 1 + size(node.right));
    }
 
    // Function to check if the permutation
    // of the sequence form Arithmetic Progression
    static boolean checkIsAP(double[] arr, int n) {
        // If the sequence contains
        // only one element
        if (n == 1)
            return true;
 
        // Sorting the array
        Arrays.sort(arr);
 
        double d = arr[1] - arr[0];
 
        // Loop to check if the sequence
        // have same common difference
        // between its consecutive elements
        for (int i = 2; i < n; i++)
            if (arr[i] - arr[i - 1] != d)
                return false;
 
        return true;
    }
 
    // Function to check if the permutation
    // of the sequence form
    // Geometric progression
    static boolean checkIsGP(double[] arr, int n) {
        // Condition when the length
        // of the sequence is 1
        if (n == 1)
            return true;
 
        // Sorting the array
        Arrays.sort(arr);
        double r = arr[1] / arr[0];
 
        // Loop to check if the the
        // sequence have same common
        // ratio in consecutive elements
        for (int i = 2; i < n; i++) {
            if (arr[i] / arr[i - 1] != r)
                return false;
        }
        return true;
    }
 
    // Function to check if the permutation
    // of the sequence form
    // Harmonic Progression
    static boolean checkIsHP(double[] arr, int n) {
        // Condition when length of
        // sequence in 1
        if (n == 1) {
            return true;
        }
        double[] rec = new double[n];
 
        // Loop to find the reciprocal
        // of the sequence
        for (int i = 0; i < n; i++) {
            rec[i] = ((1 / arr[i]));
        }
 
        // Sorting the array
        Arrays.sort(rec);
        double d = (rec[1]) - (rec[0]);
 
        // Loop to check if the common
        // difference of the sequence is same
        for (int i = 2; i < n; i++) {
            if (rec[i] - rec[i - 1] != d) {
                return false;
            }
        }
        return true;
    }
 
    // Function to check if the nodes
    // of the Binary tree forms AP/GP/HP
    static void checktype(Node root) {
 
        int n = size(root);
        double[] arr = new double[n];
        int i = 0;
 
        // Base Case
        if (root == null)
            return;
 
        // Create an empty queue
        // for level order traversal
        Queue q = new LinkedList<>();
 
        // Enqueue Root and initialize height
        q.add(root);
 
        // Loop to traverse the tree using
        // Level order Traversal
        while (q.isEmpty() == false) {
            Node node = q.poll();
            arr[i] = node.data;
            i++;
 
            // Enqueue left child
            if (node.left != null)
                q.add(node.left);
 
            // Enqueue right child
            if (node.right != null)
                q.add(node.right);
        }
 
        int flag = 0;
 
        // Condition to check if the
        // sequence form Arithmetic Progression
        if (checkIsAP(arr, n)) {
            System.out.println("Arithmetic Progression");
            flag = 1;
        }
 
        // Condition to check if the
        // sequence form Geometric Progression
        else if (checkIsGP(arr, n)) {
            System.out.println("Geometric Progression");
            flag = 1;
        }
 
        // Condition to check if the
        // sequence form Geometric Progression
        else if (checkIsHP(arr, n)) {
            System.out.println("Geometric Progression");
            flag = 1;
        } else if (flag == 0) {
            System.out.println("No");
        }
    }
 
    // Driver Code
    public static void main(String[] args) {
 
        /* Constructed Binary tree is:
                 1
                / \
               2   3
              / \   \
             4   5   8
                    / \
                   6   7
        */
        Node root = new Node(1);
        root.left = new Node(2);
        root.right = new Node(3);
        root.left.left = new Node(4);
        root.left.right = new Node(5);
        root.right.right = new Node(8);
        root.right.right.left = new Node(6);
        root.right.right.right = new Node(7);
 
        checktype(root);
    }
}
 
// This code is contributed by
// sanjeev2552

C#

// C# implementation to check that
// nodes of binary tree form AP/GP/HP
using System;
using System.Collections.Generic;
 
public class GFG {
    // Structure of the
    // node of the binary tree
    public class Node {
        public int data;
        public Node left, right;
  
        public Node(int data) {
            this.data = data;
            this.left = this.right = null;
        }
    };
  
    // Function to find the size
    // of the Binary Tree
    static int size(Node node) {
        // Base Case
        if (node == null)
            return 0;
        else
            return (size(node.left) + 1 + size(node.right));
    }
  
    // Function to check if the permutation
    // of the sequence form Arithmetic Progression
    static bool checkIsAP(double[] arr, int n) {
        // If the sequence contains
        // only one element
        if (n == 1)
            return true;
  
        // Sorting the array
        Array.Sort(arr);
  
        double d = arr[1] - arr[0];
  
        // Loop to check if the sequence
        // have same common difference
        // between its consecutive elements
        for (int i = 2; i < n; i++)
            if (arr[i] - arr[i - 1] != d)
                return false;
  
        return true;
    }
  
    // Function to check if the permutation
    // of the sequence form
    // Geometric progression
    static bool checkIsGP(double[] arr, int n) {
        // Condition when the length
        // of the sequence is 1
        if (n == 1)
            return true;
  
        // Sorting the array
        Array.Sort(arr);
        double r = arr[1] / arr[0];
  
        // Loop to check if the the
        // sequence have same common
        // ratio in consecutive elements
        for (int i = 2; i < n; i++) {
            if (arr[i] / arr[i - 1] != r)
                return false;
        }
        return true;
    }
  
    // Function to check if the permutation
    // of the sequence form
    // Harmonic Progression
    static bool checkIsHP(double[] arr, int n) {
        // Condition when length of
        // sequence in 1
        if (n == 1) {
            return true;
        }
        double[] rec = new double[n];
  
        // Loop to find the reciprocal
        // of the sequence
        for (int i = 0; i < n; i++) {
            rec[i] = ((1 / arr[i]));
        }
  
        // Sorting the array
        Array.Sort(rec);
        double d = (rec[1]) - (rec[0]);
  
        // Loop to check if the common
        // difference of the sequence is same
        for (int i = 2; i < n; i++) {
            if (rec[i] - rec[i - 1] != d) {
                return false;
            }
        }
        return true;
    }
  
    // Function to check if the nodes
    // of the Binary tree forms AP/GP/HP
    static void checktype(Node root) {
  
        int n = size(root);
        double[] arr = new double[n];
        int i = 0;
  
        // Base Case
        if (root == null)
            return;
  
        // Create an empty queue
        // for level order traversal
        Queue q = new Queue();
  
        // Enqueue Root and initialize height
        q.Enqueue(root);
  
        // Loop to traverse the tree using
        // Level order Traversal
        while (q.Count!=0 == false) {
            Node node = q.Dequeue();
            arr[i] = node.data;
            i++;
  
            // Enqueue left child
            if (node.left != null)
                q.Enqueue(node.left);
  
            // Enqueue right child
            if (node.right != null)
                q.Enqueue(node.right);
        }
  
        int flag = 0;
  
        // Condition to check if the
        // sequence form Arithmetic Progression
        if (checkIsAP(arr, n)) {
            Console.WriteLine("Arithmetic Progression");
            flag = 1;
        }
  
        // Condition to check if the
        // sequence form Geometric Progression
        else if (checkIsGP(arr, n)) {
            Console.WriteLine("Geometric Progression");
            flag = 1;
        }
  
        // Condition to check if the
        // sequence form Geometric Progression
        else if (checkIsHP(arr, n)) {
            Console.WriteLine("Geometric Progression");
            flag = 1;
        } else if (flag == 0) {
            Console.WriteLine("No");
        }
    }
  
    // Driver Code
    public static void Main(String[] args) {
  
        /* Constructed Binary tree is:
                 1
                / \
               2   3
              / \   \
             4   5   8
                    / \
                   6   7
        */
        Node root = new Node(1);
        root.left = new Node(2);
        root.right = new Node(3);
        root.left.left = new Node(4);
        root.left.right = new Node(5);
        root.right.right = new Node(8);
        root.right.right.left = new Node(6);
        root.right.right.right = new Node(7);
  
        checktype(root);
    }
}
// This code contributed by sapnasingh4991

输出：

Arithmetic Progression

性能分析：

• 时间复杂度：与上述方法一样，遍历节点并对它们进行排序，在最坏的情况下需要 O(N*logN) 时间。因此时间复杂度将为O(N*logN) 。
• 辅助空间复杂度：与上述方法一样，有额外的空间用于存储节点的数据。因此辅助空间复杂度将为O(N) 。

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