原文:
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 之间。我们需要从根开始寻找 k 级元素的乘积。根在 0 级。 注:树以下列形式给出:(节点值(左子树)(右子树)) 例:
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
进场:
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 语言(一种计算机语言，尤用于创建网站)
// 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.
python 3
# 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
服务器端编程语言（professional hypertext preprocessor 的缩写）
java 描述语言
输出:
72
时间复杂度: o(n)
https://youtu.be/y
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