表示为字符串的树中第 k 层节点的总和

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 层的元素总和。根在级别 0。
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 : 14
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 = 14 


Input : tree = "(8(3(2()())(6(5()())()))(5(10()())(7(13()())())))" 
        k = 3
Output : 9
Elements at level k = 3 are 5, 1 and 3
sum of digits of these elements = 5+1+3 = 9

1. Input 'tree' in string format and level k
2. Initialize level = -1 and sum = 0
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
           sum = sum + (ch-'0')
4. Print sum

C++

// C++ implementation to find sum of
// digits of elements at k-th level
#include 
using namespace std;
 
// Function to find sum of digits
// of elements at k-th level
int sumAtKthLevel(string tree, int k)
{
    int level = -1;
    int sum = 0;  // Initialize result
    int n = tree.length();
 
    for (int i=0; i

---

Java

// Java implementation to find sum of
// digits of elements at k-th level
class GfG {
 
// Function to find sum of digits
// of elements at k-th level
static int sumAtKthLevel(String tree, int k)
{
    int level = -1;
    int sum = 0; // Initialize result
    int n = tree.length();
 
    for (int i=0; i

---

Python3

# Python3 implementation to find sum of
# digits of elements at k-th level
 
# Function to find sum of digits
# of elements at k-th level
def sumAtKthLevel(tree, k) :
 
    level = -1
    sum = 0 # Initialize result
    n = len(tree)
 
    for i in range(n):
         
        # increasing level number
        if (tree[i] == '(') :
            level += 1
 
        # decreasing level number
        else if (tree[i] == ')'):
            level -= 1
 
        else:
         
            # check if current level is
            # the desired level or not
            if (level == k) :
                sum += (ord(tree[i]) - ord('0'))
         
    # required sum
    return sum
 
# Driver Code
if __name__ == '__main__':
    tree = "(0(5(6()())(4()(9()())))(7(1()())(3()())))"
    k = 2
    print(sumAtKthLevel(tree, k))
 
# This code is contributed by
# Shubham Singh(SHUBHAMSINGH10)

C#

// C# implementation to find sum of
// digits of elements at k-th level
 
using System;
class GfG {
 
// Function to find sum of digits
// of elements at k-th level
static int sumAtKthLevel(string tree, int k)
{
    int level = -1;
    int sum = 0; // 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)
                sum += (tree[i]-'0');
        }
    }
 
    // required sum
    return sum;
}
 
// Driver code
public static void Main()
{
    string tree = "(0(5(6()())(4()(9()())))(7(1()())(3()())))";
    int k = 2;
    Console.Write(sumAtKthLevel(tree, k));
}
}
 
// This code is contributed by Ita_c

Javascript

 
// Javascript implementation to find sum of
// digits of elements at k-th level
     
// Function to find sum of digits
// of elements at k-th level
function sumAtKthLevel(tree, k)
{
    let level = -1;
     
    // Initialize result
    let sum = 0;
    let n = tree.length;
     
    for(let 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)
                sum += (tree[i] - '0');
        }
    }
     
    // Required sum
    return sum;
}
 
// Driver code
let tree = "(0(5(6()())(4()(9()())))(7(1()())(3()())))";
let k = 2;
 
document.write(sumAtKthLevel(tree, k));
 
// This code is contributed by avanitrachhadiya2155
 

输出：

14

时间复杂度： O(n)

递归方法：想法是将字符串视为树而不实际创建树，并简单地以 Postorder Fashion 递归遍历字符串并仅考虑位于第 k 层的节点。

以下是相同的实现：

C++

// C++ implementation to find sum of
// digits of elements at k-th level
#include 
using namespace std;
 
// Recursive Function to find sum of digits
// of elements at k-th level
int sumAtKthLevel(string tree, int k,int &i,int level)
{
        
    if(tree[i++]=='(')
    {
       
      // if subtree is null, just like if root == NULL
      if(tree[i] == ')')
           return 0;           
     
      int sum=0;
       
      // Consider only level k node to be part of the sum
      if(level == k)
        sum = tree[i]-'0';
       
      // Recur for Left Subtree
      int leftsum = sumAtKthLevel(tree,k,++i,level+1);
       
      // Recur for Right Subtree
      int rightsum = sumAtKthLevel(tree,k,++i,level+1);
       
      // Taking care of ')' after left and right subtree
      ++i;
      return sum+leftsum+rightsum;       
    }
}
 
// Driver program to test above
int main()
{
    string tree = "(0(5(6()())(4()(9()())))(7(1()())(3()())))";
    int k = 2;
        int i=0;
    cout << sumAtKthLevel(tree, k,i,0);
    return 0;
}

Java

// Java implementation to find sum of
// digits of elements at k-th level
class GFG
{
    static int i;
 
    // Recursive Function to find sum of digits
    // of elements at k-th level
    static int sumAtKthLevel(String tree, int k, int level)
    {
 
        if (tree.charAt(i++) == '(')
        {
 
            // if subtree is null, just like if root == null
            if (tree.charAt(i) == ')')
                return 0;
 
            int sum = 0;
 
            // Consider only level k node to be part of the sum
            if (level == k)
                sum = tree.charAt(i) - '0';
 
            // Recur for Left Subtree
            ++i;
            int leftsum = sumAtKthLevel(tree, k, level + 1);
 
            // Recur for Right Subtree
            ++i;
            int rightsum = sumAtKthLevel(tree, k, level + 1);
 
            // Taking care of ')' after left and right subtree
            ++i;
            return sum + leftsum + rightsum;
        }
        return Integer.MIN_VALUE;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        String tree = "(0(5(6()())(4()(9()())))(7(1()())(3()())))";
        int k = 2;
        i = 0;
        System.out.print(sumAtKthLevel(tree, k, 0));
    }
}
 
// This code is contributed by 29AjayKumar

Python

# Python implementation to find sum of
# digits of elements at k-th level
 
# Recursive Function to find sum of digits
# of elements at k-th level
def sumAtKthLevel(tree, k, i, level):
     
    if(tree[i[0]] == '('):
        i[0] += 1
         
        # if subtree is null, just like if root == NULL
        if(tree[i[0]] == ')'):
            return 0           
         
        sum = 0
         
        # Consider only level k node to be part of the sum
        if(level == k):
            sum = int(tree[i[0]])
             
        # Recur for Left Subtree
        i[0] += 1
        leftsum = sumAtKthLevel(tree, k, i, level + 1)
             
        # Recur for Right Subtree
        i[0] += 1
        rightsum = sumAtKthLevel(tree, k, i, level + 1)
             
        # Taking care of ')' after left and right subtree
        i[0] += 1
        return sum + leftsum + rightsum    
     
# Driver program to test above
tree = "(0(5(6()())(4()(9()())))(7(1()())(3()())))"
k = 2
i = [0]
print(sumAtKthLevel(tree, k, i, 0))
 
# This code is contributed by SHUBHAMSINGH10

C＃

// C# implementation to find sum of
// digits of elements at k-th level
using System;
 
class GFG
{
    static int i;
 
    // Recursive Function to find sum of digits
    // of elements at k-th level
    static int sumAtKthLevel(String tree, int k, int level)
    {
 
        if (tree[i++] == '(')
        {
 
            // if subtree is null, just like if root == null
            if (tree[i] == ')')
                return 0;
 
            int sum = 0;
 
            // Consider only level k node to be part of the sum
            if (level == k)
                sum = tree[i] - '0';
 
            // Recur for Left Subtree
            ++i;
            int leftsum = sumAtKthLevel(tree, k, level + 1);
 
            // Recur for Right Subtree
            ++i;
            int rightsum = sumAtKthLevel(tree, k, level + 1);
 
            // Taking care of ')' after left and right subtree
            ++i;
            return sum + leftsum + rightsum;
        }
        return int.MinValue;
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        String tree = "(0(5(6()())(4()(9()())))(7(1()())(3()())))";
        int k = 2;
        i = 0;
        Console.Write(sumAtKthLevel(tree, k, 0));
    }
}
 
// This code is contributed by Rajput-Ji

Javascript

// Javascript implementation to find sum of
// digits of elements at k-th level
     
    // Recursive Function to find sum of digits
    // of elements at k-th level
    function sumAtKthLevel(tree,k,level)
    {
        if (tree[i++] == '(')
        {
  
            // if subtree is null, just like if root == null
            if (tree[i] == ')')
                return 0;
  
            let sum = 0;
  
            // Consider only level k node to be part of the sum
            if (level == k)
                sum = tree[i] - '0';
  
            // Recur for Left Subtree
            ++i;
            let leftsum = sumAtKthLevel(tree, k, level + 1);
  
            // Recur for Right Subtree
            ++i;
            let rightsum = sumAtKthLevel(tree, k, level + 1);
  
            // Taking care of ')' after left and right subtree
            ++i;
            return sum + leftsum + rightsum;
        }
        return Number.MIN_VALUE;
    }
     
     // Driver code
    let tree = "(0(5(6()())(4()(9()())))(7(1()())(3()())))";
    let k = 2;
    let i = 0;
    document.write(sumAtKthLevel(tree, k, 0));
     
    // This code is contributed by rag2127

输出 ：

14

时间复杂度： O(n)