仅使用单个递归函数对总计为给定值x的子树进行计数

给定一个包含n个节点的二叉树。问题是仅使用单个递归函数对具有总节点数据总和等于给定值的子树进行计数。

例子：

Input : 
             5
           /   \  
        -10     3
        /  \   /  \
       9    8 -4   7
       
       x = 7

Output : 2
There are 2 subtrees with sum 7.

1st one is leaf node:
7.

2nd one is:

      -10
     /   \
    9     8

资料来源： Microsoft面试体验|设置157。

方法：

countSubtreesWithSumX(root, count, x)
    if !root then
        return 0
        
    ls = countSubtreesWithSumX(root->left, count, x)
    rs = countSubtreesWithSumX(root->right, count, x)
    sum = ls + rs + root->data
    
    if sum == x then
    count++
    return sum

countSubtreesWithSumXUtil(root, x)
    if !root then
        return 0
    
    Initialize count = 0
    ls = countSubtreesWithSumX(root->left, count, x)
    rs = countSubtreesWithSumX(root->right, count, x)
    
    if (ls + rs + root->data) == x
        count++
    return count

C++

// C++ implementation to count subtress that
// sum up to a given value x
#include 
 
using namespace std;
 
// structure of a node of binary tree
struct Node {
    int data;
    Node *left, *right;
};
 
// function to get a new node
Node* getNode(int data)
{
    // allocate space
    Node* newNode = (Node*)malloc(sizeof(Node));
 
    // put in the data
    newNode->data = data;
    newNode->left = newNode->right = NULL;
    return newNode;
}
 
// function to count subtress that
// sum up to a given value x
int countSubtreesWithSumX(Node* root,
                          int& count, int x)
{
    // if tree is empty
    if (!root)
        return 0;
 
    // sum of nodes in the left subtree
    int ls = countSubtreesWithSumX(root->left, count, x);
 
    // sum of nodes in the right subtree
    int rs = countSubtreesWithSumX(root->right, count, x);
 
    // sum of nodes in the subtree rooted
    // with 'root->data'
    int sum = ls + rs + root->data;
 
    // if true
    if (sum == x)
        count++;
 
    // return subtree's nodes sum
    return sum;
}
 
// utility function to count subtress that
// sum up to a given value x
int countSubtreesWithSumXUtil(Node* root, int x)
{
    // if tree is empty
    if (!root)
        return 0;
 
    int count = 0;
 
    // sum of nodes in the left subtree
    int ls = countSubtreesWithSumX(root->left, count, x);
 
    // sum of nodes in the right subtree
    int rs = countSubtreesWithSumX(root->right, count, x);
 
    // if tree's nodes sum == x
    if ((ls + rs + root->data) == x)
        count++;
 
    // required count of subtrees
    return count;
}
 
// Driver program to test above
int main()
{
    /* binary tree creation   
                5
              /   \ 
           -10     3
           /  \   /  \
          9    8 -4   7
    */
    Node* root = getNode(5);
    root->left = getNode(-10);
    root->right = getNode(3);
    root->left->left = getNode(9);
    root->left->right = getNode(8);
    root->right->left = getNode(-4);
    root->right->right = getNode(7);
 
    int x = 7;
 
    cout << "Count = "
         << countSubtreesWithSumXUtil(root, x);
 
    return 0;
}

Java

// Java program to find if
// there is a subtree with
// given sum
import java.util.*;
class GFG
{
 
// structure of a node
// of binary tree
static class Node
{
    int data;
    Node left, right;
}
 
static class INT
{
    int v;
    INT(int a)
    {
        v = a;
    }
}
 
// function to get a new node
static Node getNode(int data)
{
    // allocate space
    Node newNode = new Node();
 
    // put in the data
    newNode.data = data;
    newNode.left = newNode.right = null;
    return newNode;
}
 
// function to count subtress that
// sum up to a given value x
static int countSubtreesWithSumX(Node root,
                          INT count, int x)
{
    // if tree is empty
    if (root == null)
        return 0;
 
    // sum of nodes in the left subtree
    int ls = countSubtreesWithSumX(root.left,
                                   count, x);
 
    // sum of nodes in the right subtree
    int rs = countSubtreesWithSumX(root.right,
                                   count, x);
 
    // sum of nodes in the subtree
    // rooted with 'root.data'
    int sum = ls + rs + root.data;
 
    // if true
    if (sum == x)
        count.v++;
 
    // return subtree's nodes sum
    return sum;
}
 
// utility function to
// count subtress that
// sum up to a given value x
static int countSubtreesWithSumXUtil(Node root,
                                     int x)
{
    // if tree is empty
    if (root == null)
        return 0;
 
    INT count = new INT(0);
 
    // sum of nodes in the left subtree
    int ls = countSubtreesWithSumX(root.left,
                                   count, x);
 
    // sum of nodes in the right subtree
    int rs = countSubtreesWithSumX(root.right,
                                   count, x);
 
    // if tree's nodes sum == x
    if ((ls + rs + root.data) == x)
        count.v++;
 
    // required count of subtrees
    return count.v;
}
 
// Driver Code
public static void main(String args[])
{
    /* binary tree creation    
                5
            / \
        -10     3
        / \ / \
        9 8 -4 7
    */
    Node root = getNode(5);
    root.left = getNode(-10);
    root.right = getNode(3);
    root.left.left = getNode(9);
    root.left.right = getNode(8);
    root.right.left = getNode(-4);
    root.right.right = getNode(7);
 
    int x = 7;
 
    System.out.println("Count = " +
           countSubtreesWithSumXUtil(root, x));
}
}
 
// This code is contributed
// by Arnab Kundu

Python3

# Python3 implementation to count subtress
# that Sum up to a given value x
 
# class to get a new node
class getNode:
    def __init__(self, data):
         
        # put in the data
        self.data = data
        self.left = self.right = None
         
# function to count subtress that
# Sum up to a given value x
def countSubtreesWithSumX(root, count, x):
     
    # if tree is empty
    if (not root):
        return 0
 
    # Sum of nodes in the left subtree
    ls = countSubtreesWithSumX(root.left,
                               count, x)
 
    # Sum of nodes in the right subtree
    rs = countSubtreesWithSumX(root.right,
                               count, x)
 
    # Sum of nodes in the subtree
    # rooted with 'root.data'
    Sum = ls + rs + root.data
 
    # if true
    if (Sum == x):
        count[0] += 1
 
    # return subtree's nodes Sum
    return Sum
 
# utility function to count subtress
# that Sum up to a given value x
def countSubtreesWithSumXUtil(root, x):
     
    # if tree is empty
    if (not root):
        return 0
 
    count = [0]
 
    # Sum of nodes in the left subtree
    ls = countSubtreesWithSumX(root.left,
                               count, x)
 
    # Sum of nodes in the right subtree
    rs = countSubtreesWithSumX(root.right,
                               count, x)
 
    # if tree's nodes Sum == x
    if ((ls + rs + root.data) == x):
        count[0] += 1
 
    # required count of subtrees
    return count[0]
 
# Driver Code
if __name__ == '__main__':
     
    # binary tree creation    
    #         5
    #         / \
    #     -10     3
    #     / \ / \
    #     9 8 -4 7
    root = getNode(5)
    root.left = getNode(-10)
    root.right = getNode(3)
    root.left.left = getNode(9)
    root.left.right = getNode(8)
    root.right.left = getNode(-4)
    root.right.right = getNode(7)
 
    x = 7
 
    print("Count =",
           countSubtreesWithSumXUtil(root, x))
 
# This code is contributed by PranchalK

C#

using System;
 
// c# program to find if
// there is a subtree with
// given sum
public class GFG
{
 
// structure of a node
// of binary tree
public class Node
{
    public int data;
    public Node left, right;
}
 
public class INT
{
    public int v;
    public INT(int a)
    {
        v = a;
    }
}
 
// function to get a new node
public static Node getNode(int data)
{
    // allocate space
    Node newNode = new Node();
 
    // put in the data
    newNode.data = data;
    newNode.left = newNode.right = null;
    return newNode;
}
 
// function to count subtress that
// sum up to a given value x
public static int countSubtreesWithSumX(Node root,
                                    INT count, int x)
{
    // if tree is empty
    if (root == null)
    {
        return 0;
    }
 
    // sum of nodes in the left subtree
    int ls = countSubtreesWithSumX(root.left, count, x);
 
    // sum of nodes in the right subtree
    int rs = countSubtreesWithSumX(root.right, count, x);
 
    // sum of nodes in the subtree
    // rooted with 'root.data'
    int sum = ls + rs + root.data;
 
    // if true
    if (sum == x)
    {
        count.v++;
    }
 
    // return subtree's nodes sum
    return sum;
}
 
// utility function to
// count subtress that
// sum up to a given value x
public static int countSubtreesWithSumXUtil(Node root, int x)
{
    // if tree is empty
    if (root == null)
    {
        return 0;
    }
 
    INT count = new INT(0);
 
    // sum of nodes in the left subtree
    int ls = countSubtreesWithSumX(root.left, count, x);
 
    // sum of nodes in the right subtree
    int rs = countSubtreesWithSumX(root.right, count, x);
 
    // if tree's nodes sum == x
    if ((ls + rs + root.data) == x)
    {
        count.v++;
    }
 
    // required count of subtrees
    return count.v;
}
 
// Driver Code
public static void Main(string[] args)
{
    /* binary tree creation    
                5
            / \
        -10     3
        / \ / \
        9 8 -4 7
    */
    Node root = getNode(5);
    root.left = getNode(-10);
    root.right = getNode(3);
    root.left.left = getNode(9);
    root.left.right = getNode(8);
    root.right.left = getNode(-4);
    root.right.right = getNode(7);
 
    int x = 7;
 
    Console.WriteLine("Count = " + countSubtreesWithSumXUtil(root, x));
}
}
 
// This code is contributed by Shrikant13

输出：

Count = 2

时间复杂度： O(n)。

另一种方法：

countSubtreesWithSumXUtil(root, x)

 Initialize static count = 0
 Initialize static *ptr = root
    if !root then
        return 0
    
    Initialize static count = 0
    ls+ = countSubtreesWithSumXUtil(root->left, count, x)
    rs+ = countSubtreesWithSumXUtil(root->right, count, x)
    
    if (ls + rs + root->data) == x
        count++
    
    if(ptr!=root)
       return ls + root->data + rs
    else
    return count

C++

// C++ program to find if
// there is a subtree with
// given sum
#include 
 
using namespace std;
 
// Structure of a node of binary tree
struct Node {
    int data;
    Node *left, *right;
};
 
// Function to get a new node
Node* getNode(int data)
{
    // Allocate space
    Node* newNode = (Node*)malloc(sizeof(Node));
 
    // Put in the data
    newNode->data = data;
    newNode->left = newNode->right = NULL;
    return newNode;
}
 
 
 
 
// Utility function to count subtress that
// sum up to a given value x
int countSubtreesWithSumXUtil(Node* root, int x)
{
    static int count=0;
    static Node* ptr=root;
    int l=0,r=0;
    if(root==NULL)
    return 0;
     
    l+=countSubtreesWithSumXUtil(root->left,x);
     
    r+=countSubtreesWithSumXUtil(root->right,x);
 
    if(l+r+root->data==x)
    count++;
 
    if(ptr!=root)
        return l+root->data+r;
     
    return count;
     
         
}
 
// Driver code
int main()
{
    /* binary tree creation
              5
            /  \
          -10   3
          / \  / \
          9 8 -4 7
    */
    Node* root = getNode(5);
    root->left = getNode(-10);
    root->right = getNode(3);
    root->left->left = getNode(9);
    root->left->right = getNode(8);
    root->right->left = getNode(-4);
    root->right->right = getNode(7);
 
    int x = 7;
 
    cout << "Count = "
        << countSubtreesWithSumXUtil(root, x);
 
    return 0;
}
// This code is contributed by Sadik Ali

Java

// Java program to find if
// there is a subtree with
// given sum
import java.io.*;
 
// Node class to create new node
class Node
{
    int data;
    Node left;
    Node right;
    Node(int data)
    {
        this.data = data;
    }
}
 
class GFG
{
    static int count = 0;
    static Node ptr;
     
    // Utility function to count subtress that
    // sum up to a given value x
    int countSubtreesWithSumXUtil(Node root, int x)
    {
        int l = 0, r = 0;
        if(root == null) return 0;
        l += countSubtreesWithSumXUtil(root.left, x);
        r += countSubtreesWithSumXUtil(root.right, x);
        if(l + r + root.data == x) count++;
        if(ptr != root) return l + root.data + r;
        return count;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        /* binary tree creation
            5
        / \
        -10 3
        / \ / \
        9 8 -4 7
        */
        Node root = new Node(5);
        root.left = new Node(-10);
        root.right = new Node(3);
        root.left.left = new Node(9);
        root.left.right = new Node(8);
        root.right.left = new Node(-4);
        root.right.right = new Node(7);
        int x = 7;
        ptr = root; // assigning global value of ptr
        System.out.println("Count = " +
               new GFG().countSubtreesWithSumXUtil(root, x));
    }
}
 
// This code is submitted by Devarshi_Singh

Python3

# Python3 program to find if there is
# a subtree with given sum
  
# Structure of a node of binary tree
class Node:
     
    def __init__(self, data):
         
        self.data = data
        self.left = None
        self.right = None
  
# Function to get a new node
def getNode(data):
 
    # Allocate space
    newNode = Node(data)
    return newNode
 
count = 0
ptr = None
 
# Utility function to count subtress that
# sum up to a given value x
def countSubtreesWithSumXUtil(root, x):
     
    global count, ptr
 
    l = 0
    r = 0
     
    if (root == None):
        return 0  
      
    l += countSubtreesWithSumXUtil(root.left, x)
    r += countSubtreesWithSumXUtil(root.right, x)
  
    if (l + r + root.data == x):
        count += 1
  
    if (ptr != root):
        return l + root.data + r
      
    return count
      
# Driver code
if __name__=='__main__':
     
    ''' binary tree creation
              5
            /  \
          -10   3
          / \  / \
          9 8 -4 7
    '''
     
    root = getNode(5)
    root.left = getNode(-10)
    root.right = getNode(3)
    root.left.left = getNode(9)
    root.left.right = getNode(8)
    root.right.left = getNode(-4)
    root.right.right = getNode(7)
  
    x = 7
    ptr = root
     
    print("Count = " + str(countSubtreesWithSumXUtil(
        root, x)))
  
# This code is contributed by pratham76

C＃

// C# program to find if
// there is a subtree with
// given sum
using System;
  
// Node class to
// create new node
public class Node
{
  public int data;
  public Node left;
  public Node right;
  public Node(int data)
  {
    this.data = data;
  }
}
  
class GFG{
 
static int count = 0;
static Node ptr;
 
// Utility function to count subtress
// that sum up to a given value x
int countSubtreesWithSumXUtil(Node root,
                              int x)
{
  int l = 0, r = 0;
  if(root == null) return 0;
  l += countSubtreesWithSumXUtil(root.left, x);
  r += countSubtreesWithSumXUtil(root.right, x);
  if(l + r + root.data == x) count++;
  if(ptr != root) return l + root.data + r;
  return count;
}
 
// Driver Code
public static void Main(string[] args)
{
  /* binary tree creation
            5
        / \
        -10 3
        / \ / \
        9 8 -4 7
        */
  Node root = new Node(5);
  root.left = new Node(-10);
  root.right = new Node(3);
  root.left.left = new Node(9);
  root.left.right = new Node(8);
  root.right.left = new Node(-4);
  root.right.right = new Node(7);
  int x = 7;
   
  // Assigning global value of ptr
  ptr = root;
  Console.Write("Count = " +
          new GFG().countSubtreesWithSumXUtil(root, x));
}
}
 
// This code is contributed by rutvik_56

输出：

Count = 2

时间复杂度： O(n)。