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)