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📜  二叉树中所有节点的乘积

📅  最后修改于: 2022-05-13 01:57:05.172000             🧑  作者: Mango

二叉树中所有节点的乘积

给定一棵二叉树。任务是编写一个程序来查找给定二叉树的所有节点的乘积。

在上面的二叉树中,
产品= 15*10*8*12*20*16*25 = 115200000

这个想法是递归地:

  • 求左子树的乘积。
  • 找到右子树的乘积。
  • 将左右子树的乘积与当前节点的数据相乘并返回。

下面是上述方法的实现:

C++
// Program to print product of all
// the nodes of a binary tree
#include 
using namespace std;
 
// Binary Tree Node
struct Node {
    int key;
    Node *left, *right;
};
 
/* utility that allocates a new Node
   with the given key */
Node* newNode(int key)
{
    Node* node = new Node;
    node->key = key;
    node->left = node->right = NULL;
    return (node);
}
 
// Function to find product of
// all the nodes
int productBT(Node* root)
{
    if (root == NULL)
        return 1;
 
    return (root->key * productBT(root->left) * productBT(root->right));
}
 
// Driver Code
int main()
{
    // Binary Tree is:
    //       1
    //      /  \
    //     2    3
    //    / \  / \
    //   4   5 6  7
    //          \
    //           8
    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);
 
    int prod = productBT(root);
 
    cout << "Product of all the nodes is: "
         << prod << endl;
 
    return 0;
}


Java
// Java Program to print product of all
// the nodes of a binary tree
import java.util.*;
class solution
{
 
// Binary Tree Node
static class Node {
    int key;
    Node left, right;
};
 
/* utility that allocates a new Node
   with the given key */
static Node newNode(int key)
{
    Node node = new Node();
    node.key = key;
    node.left = node.right = null;
    return (node);
}
 
// Function to find product of
// all the nodes
static int productBT(Node root)
{
    if (root == null)
        return 1;
 
    return (root.key * productBT(root.left) * productBT(root.right));
}
 
// Driver Code
public static void main(String args[])
{
    // Binary Tree is:
    //       1
    //      /  \
    //     2    3
    //    / \  / \
    //   4   5 6  7
    //          \
    //           8
    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);
 
    int prod = productBT(root);
 
    System.out.println( "Product of all the nodes is: "+prod);
 
}
}
//contributed by Arnab Kundu


Python3
# Python3 Program to print product of
# all the nodes of a binary tree
 
# Binary Tree Node
 
""" utility that allocates a new Node
with the given key """
class newNode:
 
    # Construct to create a new node
    def __init__(self, key):
        self.key = key
        self.left = None
        self.right = None
         
# Function to find product of
# all the nodes
def productBT( root) :
 
    if (root == None):
        return 1
 
    return (root.key * productBT(root.left) *
                       productBT(root.right))
 
# Driver Code
if __name__ == '__main__':
     
    # Binary Tree is:
    #     1
    #     / \
    #     2 3
    # / \ / \
    # 4 5 6 7
    #         \
    #         8
    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)
 
    prod = productBT(root)
 
    print("Product of all the nodes is:", prod)
     
# This code is contributed by
# Shubham Singh(SHUBHAMSINGH10)


C#
// C# Program to print product of all
// the nodes of a binary tree
using System;
 
class GFG
{
 
    // Binary Tree Node
    class Node
    {
        public int key;
        public Node left, right;
    };
 
    /* utility that allocates a new Node
    with the given key */
    static Node newNode(int key)
    {
        Node node = new Node();
        node.key = key;
        node.left = node.right = null;
        return (node);
    }
 
    // Function to find product of
    // all the nodes
    static int productBT(Node root)
    {
        if (root == null)
            return 1;
 
        return (root.key * productBT(root.left) *
                        productBT(root.right));
    }
 
    // Driver Code
    public static void Main()
    {
        // Binary Tree is:
        //     1
        //     / \
        //     2 3
        // / \ / \
        // 4 5 6 7
        //         \
        //         8
        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);
 
        int prod = productBT(root);
 
        Console.WriteLine( "Product of all " +
                        "the nodes is: " + prod);
    }
}
 
/* This code is contributed PrinciRaj1992 */


Javascript


Javascript


输出:

Product of all the nodes is: 40320

时间复杂度: O(n)