表示为字符串的树中第 k 级节点的乘积

给定一个整数“K”和一个字符串格式的二叉树。树的每个节点的值都在 0 到 9 的范围内。我们需要从根开始找到第 K 层元素的乘积。根在级别 0。
注意：树以以下形式给出：（节点值（左子树）（右子树））
例子：

Input : tree = "(0(5(6()())(4()(9()())))(7(1()())(3()())))" 
        k = 2
Output : 72
Its tree representation is shown below

Elements at level k = 2 are 6, 4, 1, 3
sum of the digits of these elements = 6 * 4 * 1 * 3 = 72 


Input : tree = "(8(3(2()())(6(5()())()))(5(10()())(7(13()())())))" 
        k = 3
Output : 15
Elements at level k = 3 are 5, 1 and 3
sum of digits of these elements = 5 * 1 * 3 = 15

方法：

1. Input 'tree' in string format and level k
2. Initialize level = -1 and product = 1
3. for each character 'ch' in 'tree'
   3.1  if ch == '(' then
        --> level++
   3.2  else if ch == ')' then
        --> level--
   3.3  else
        if level == k then
           product = product * (ch-'0')
4. Print product

C++

// C++ implementation to find product of
// digits of elements at k-th level
#include 
using namespace std;
 
// Function to find product of digits
// of elements at k-th level
int productAtKthLevel(string tree, int k)
{
    int level = -1;
    int product = 1; // Initialize result
    int n = tree.length();
 
    for (int i = 0; i < n; i++) {
        // increasing level number
        if (tree[i] == '(')
            level++;
 
        // decreasing level number
        else if (tree[i] == ')')
            level--;
 
        else {
            // check if current level is
            // the desired level or not
            if (level == k)
                product *= (tree[i] - '0');
        }
    }
 
    // required product
    return product;
}
 
// Driver program
int main()
{
    string tree = "(0(5(6()())(4()(9()())))(7(1()())(3()())))";
    int k = 2;
    cout << productAtKthLevel(tree, k);
    return 0;
}

Java

// Java implementation to find product of
// digits of elements at k-th level
 
class GFG
{
    // Function to find product of digits
    // of elements at k-th level
    static int productAtKthLevel(String tree, int k)
    {
        int level = -1;
         
        // Initialize result
        int product = 1;
         
        int n = tree.length();
     
        for (int i = 0; i < n; i++)
        {
            // increasing level number
            if (tree.charAt(i) == '(')
                level++;
     
            // decreasing level number
            else if (tree.charAt(i) == ')')
                level--;
     
            else
            {
                // check if current level is
                // the desired level or not
                if (level == k)
                    product *= (tree.charAt(i) - '0');
            }
        }
     
        // required product
        return product;
    }
     
    // Driver program
    public static void main(String[] args)
    {
        String tree = "(0(5(6()())(4()(9()())))(7(1()())(3()())))";
        int k = 2;
        System.out.println(productAtKthLevel(tree, k));
    }
}
 
// This code is contributed
// by Smitha Dinesh Semwal.

Python3

# Python 3 implementation
# to find product of
# digits of elements
# at k-th level
 
# Function to find
# product of digits
# of elements at
# k-th level
def productAtKthLevel(tree, k):
 
    level = -1
         
        # Initialize result
    product = 1
    n = len(tree)
 
    for i in range(0, n):
 
        # increasing level number
        if (tree[i] == '('):
            level+=1
 
        # decreasing level number
        elif (tree[i] == ')'):
            level-=1
 
        else:
            # check if current level is
            # the desired level or not
            if (level == k):
                product *= (int(tree[i]) - int('0'))
         
     
 
    # required product
    return product
 
 
# Driver program
tree = "(0(5(6()())(4()(9()())))(7(1()())(3()())))"
k = 2
 
print(productAtKthLevel(tree, k))
 
# This code is contributed by
# Smitha Dinesh Semwal

C#

// C# implementation to find
// product of digits of
// elements at k-th level
using System;
 
class GFG
{
    // Function to find product
    // of digits of elements
    // at k-th level
    static int productAtKthLevel(string tree,
                                 int k)
    {
        int level = -1;
         
        // Initialize result
        int product = 1;
         
        int n = tree.Length;
     
        for (int i = 0; i < n; i++)
        {
            // increasing
            // level number
            if (tree[i] == '(')
                level++;
     
            // decreasing
            // level number
            else if (tree[i] == ')')
                level--;
     
            else
            {
                // check if current level is
                // the desired level or not
                if (level == k)
                    product *= (tree[i] - '0');
            }
        }
     
        // required product
        return product;
    }
     
    // Driver Code
    static void Main()
    {
        string tree = "(0(5(6()())(4()(9()())))(7(1()())(3()())))";
        int k = 2;
        Console.WriteLine(productAtKthLevel(tree, k));
    }
}
 
// This code is contributed by Sam007

PHP

Javascript

输出：

时间复杂度： O(n)

https://youtu.be/Y

-yARmXlSbQ?list=PLqM7alHXFySHCXD7r1J0ky9Zg_GBB1dbk