给定二叉树中特殊平衡节点的总和

给定二叉树，任务是在给定的二叉树中找到所有特殊平衡节点的总和。

A specially balanced node in a Binary Tree contains the sum of nodes of one subtree(either left or right) as even and the sum of the other subtree as odd.
The nodes having only one child or no child can never be a balanced node.

例子：

Input: Below is the given Tree: 
 

Output: 33
Explanation:
The specially balanced nodes are 11 and 22.
For Node 11:
The sum of left subtree is (23 + 13 + 9) = 45 which is Odd.
The sum of right subtree is (44 + 22 + 7 + 6 + 15) = 94 which is Even.
For Node 22:
The sum of left subtree is 6 which is Even.
The sum of right subtree is 15 which is Odd.
Therefore, the sum of specially balanced nodes is 11 + 22 = 33.

Input: Below is the given Tree: 
 

Output: 16
Explanation:
The specially balanced nodes are 4 and 12.
For Node 4:
The sum of left subtree is 2 which is Even.
The sum of right subtree is 3 which is Odd.
For Node 12:
The sum of left subtree is 17 which is Odd.
The sum of right subtree is (16 + 4 + 9 + 2 + 3) = 34 which is Even.
Therefore, the sum of specially balanced nodes is 4 + 12 = 16.

方法：想法是使用递归在给定的Tree上执行DFS遍历，并根据给定的条件更新最终的总和。请按照以下步骤操作：

1. 将totalSum初始化为0 ，以存储所有特殊平衡的节点之和。
2. 在给定的树上执行DFS遍历，并检查以下内容：
   • 如果该节点是叶节点，则返回该节点的值。
   • 现在，如果当前节点不是叶节点，则对左和右子树进行递归遍历。
   • 每次递归调用结束时，返回带有当前根值的左右子树的总和值。
   • 检查返回的和是否满足特殊平衡节点的属性。如果发现为真，则将当前节点值添加到totalSum中。
3. 完成上述步骤后，打印totalSum的值。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Structure of Binary Tree
struct Node {
    int data;
    Node *left, *right;
};
 
// Function to create a new node
Node* newnode(int data)
{
    Node* temp = new Node;
    temp->data = data;
    temp->left = NULL;
    temp->right = NULL;
 
    // Return the created node
    return temp;
}
 
// Function to insert a node in the tree
Node* insert(string s, int i, int N,
             Node* root, Node* temp)
{
    if (i == N)
        return temp;
 
    // Left insertion
    if (s[i] == 'L')
        root->left = insert(s, i + 1, N,
                            root->left,
                            temp);
 
    // Right insertion
    else
        root->right = insert(s, i + 1, N,
                             root->right,
                             temp);
 
    // Return the root node
    return root;
}
 
// Function to find sum of specially
// balanced nodes in the Tree
int SBTUtil(Node* root, int& sum)
{
    // Base Case
    if (root == NULL)
        return 0;
 
    if (root->left == NULL
        && root->right == NULL)
        return root->data;
 
    // Find the left subtree sum
    int left = SBTUtil(root->left, sum);
 
    // Find the right subtree sum
    int right = SBTUtil(root->right, sum);
 
    // Condition of specially
    // balanced node
    if (root->left && root->right) {
 
        // Condition of specially
        // balanced node
        if ((left % 2 == 0
             && right % 2 != 0)
            || (left % 2 != 0
                && right % 2 == 0)) {
 
            sum += root->data;
        }
    }
 
    // Return the sum
    return left + right + root->data;
}
 
// Function to build the binary tree
Node* build_tree(int R, int N,
                 string str[],
                 int values[])
{
    // Form root node of the tree
    Node* root = newnode(R);
    int i;
 
    // Insert nodes into tree
    for (i = 0; i < N - 1; i++) {
        string s = str[i];
        int x = values[i];
 
        // Create a new Node
        Node* temp = newnode(x);
 
        // Insert the node
        root = insert(s, 0, s.size(),
                      root, temp);
    }
 
    // Return the root of the Tree
    return root;
}
 
// Function to find the sum of specially
// balanced nodes
void speciallyBalancedNodes(
    int R, int N, string str[], int values[])
{
    // Build Tree
    Node* root = build_tree(R, N,
                            str, values);
 
    // Stores the sum of specially
    // balanced node
    int sum = 0;
 
    // Function Call
    SBTUtil(root, sum);
 
    // Print required sum
    cout << sum << " ";
}
 
// Driver Code
int main()
{
    // Given nodes
    int N = 7;
 
    // Given root
    int R = 12;
 
    // Given path info of nodes
    // from root
    string str[N - 1]
        = { "L", "R", "RL",
            "RR", "RLL", "RLR" };
 
    // Given node values
    int values[N - 1] = { 17, 16, 4,
                          9, 2, 3 };
 
    // Function Call
    speciallyBalancedNodes(R, N, str,
                           values);
 
    return 0;
}

Java

// Java program for
// the above approach
import java.util.*;
class GFG{
   
static int sum;
 
//Structure of Binary Tree
static class Node
{
  int data;
  Node left, right;
};
 
//Function to create a new node
static Node newnode(int data)
{
  Node temp = new Node();
  temp.data = data;
  temp.left = null;
  temp.right = null;
 
  // Return the created node
  return temp;
}
 
//Function to insert a node in the tree
static Node insert(String s, int i, int N,
                   Node root, Node temp)
{
  if (i == N)
    return temp;
 
  // Left insertion
  if (s.charAt(i) == 'L')
    root.left = insert(s, i + 1, N,
                       root.left, temp);
 
  // Right insertion
  else
    root.right = insert(s, i + 1, N,
                        root.right, temp);
   
  // Return the root node
  return root;
}
 
//Function to find sum of specially
//balanced nodes in the Tree
static int SBTUtil(Node root)
{
  // Base Case
  if (root == null)
    return 0;
 
  if (root.left == null &&
      root.right == null)
    return root.data;
 
  // Find the left subtree sum
  int left = SBTUtil(root.left);
 
  // Find the right subtree sum
  int right = SBTUtil(root.right);
 
  // Condition of specially
  // balanced node
  if (root.left != null &&
      root.right != null)
  {
    // Condition of specially
    // balanced node
    if ((left % 2 == 0 && right % 2 != 0) ||
        (left % 2 != 0 && right % 2 == 0))
    {
      sum += root.data;
    }
  }
 
  // Return the sum
  return left + right + root.data;
}
 
//Function to build the binary tree
static Node build_tree(int R, int N,
                       String str[],
                       int values[])
{
  // Form root node of the tree
  Node root = newnode(R);
  int i;
 
  // Insert nodes into tree
  for (i = 0; i < N - 1; i++)
  {
    String s = str[i];
    int x = values[i];
 
    // Create a new Node
    Node temp = newnode(x);
 
    // Insert the node
    root = insert(s, 0, s.length(),
                  root, temp);
  }
 
  // Return the root of the Tree
  return root;
}
 
// Function to find the
// sum of specially
// balanced nodes
static void speciallyBalancedNodes(int R, int N,
                                   String str[],
                                   int values[])
{
  // Build Tree
  Node root = build_tree(R, N, str,
                         values);
 
  // Stores the sum of specially
  // balanced node
  sum = 0;
 
  // Function Call
  SBTUtil(root);
 
  // Print required sum
  System.out.print(sum + " ");
}
 
//Driver Code
public static void main(String[] args)
{
  // Given nodes
  int N = 7;
 
  // Given root
  int R = 12;
 
  // Given path info of nodes
  // from root
  String str[] = {"L", "R", "RL",
                 "RR", "RLL", "RLR"};
 
  // Given node values
  int values[] = {17, 16, 4,
                  9, 2, 3};
 
  // Function Call
  speciallyBalancedNodes(R, N, str,
                         values);
}
}
 
//This code is contributed by 29AjayKumar

Python3

# Python3 program for the above approach
 
# Structure of Binary Tree
class Node:
 
    def __init__(self, data):
         
        self.data = data
        self.left = None
        self.right = None
 
# Function to create a new node
def newnode(data):
 
    temp = Node(data)
     
    # Return the created node
    return temp
 
# Function to insert a node in the tree
def insert(s, i, N, root, temp):
     
    if (i == N):
        return temp
 
    # Left insertion
    if (s[i] == 'L'):
        root.left = insert(s, i + 1, N,
                           root.left, temp)
 
    # Right insertion
    else:
        root.right = insert(s, i + 1, N,
                            root.right, temp)
 
    # Return the root node
    return root
 
# Function to find sum of specially
# balanced nodes in the Tree
def SBTUtil(root, sum):
     
    # Base Case
    if (root == None):
        return [0, sum]
 
    if (root.left == None and
       root.right == None):
        return [root.data, sum]
 
    # Find the left subtree sum
    left, sum = SBTUtil(root.left, sum)
 
    # Find the right subtree sum
    right, sum = SBTUtil(root.right, sum)
 
    # Condition of specially
    # balanced node
    if (root.left and root.right):
         
        # Condition of specially
        # balanced node
        if ((left % 2 == 0 and
            right % 2 != 0) or
            (left % 2 != 0 and
            right % 2 == 0)):
            sum += root.data
 
    # Return the sum
    return [left + right + root.data, sum]
 
# Function to build the binary tree
def build_tree(R, N, str, values):
     
    # Form root node of the tree
    root = newnode(R)
 
    # Insert nodes into tree
    for i in range(0, N - 1):
        s = str[i]
        x = values[i]
 
        # Create a new Node
        temp = newnode(x)
 
        # Insert the node
        root = insert(s, 0, len(s),
                      root, temp)
     
    # Return the root of the Tree
    return root
 
# Function to find the sum of specially
# balanced nodes
def speciallyBalancedNodes(R, N, str, values):
 
    # Build Tree
    root = build_tree(R, N, str, values)
 
    # Stores the sum of specially
    # balanced node
    sum = 0
 
    # Function Call
    tmp, sum = SBTUtil(root, sum)
 
    # Print required sum
    print(sum, end = ' ')
 
# Driver code
if __name__ == "__main__":
     
    # Given nodes
    N = 7
 
    # Given root
    R = 12
 
    # Given path info of nodes
    # from root
    str = [ "L", "R", "RL",
            "RR", "RLL", "RLR" ]
 
    # Given node values
    values = [ 17, 16, 4, 9, 2, 3 ]
 
    # Function Call
    speciallyBalancedNodes(R, N, str,
                           values)
 
# This code is contributed by rutvik_56

C#

//C# program for
// the above approach
using System;
class GFG{
   
static int sum;
 
//Structure of Binary Tree
class Node
{
  public int data;
  public Node left, right;
};
 
//Function to create a new node
static Node newnode(int data)
{
  Node temp = new Node();
  temp.data = data;
  temp.left = null;
  temp.right = null;
 
  // Return the created node
  return temp;
}
 
//Function to insert a node in the tree
static Node insert(String s, int i, int N,
                   Node root, Node temp)
{
  if (i == N)
    return temp;
 
  // Left insertion
  if (s[i] == 'L')
    root.left = insert(s, i + 1, N,
                       root.left, temp);
 
  // Right insertion
  else
    root.right = insert(s, i + 1, N,
                        root.right, temp);
   
  // Return the root node
  return root;
}
 
//Function to find sum of specially
//balanced nodes in the Tree
static int SBTUtil(Node root)
{
  // Base Case
  if (root == null)
    return 0;
 
  if (root.left == null &&
      root.right == null)
    return root.data;
 
  // Find the left subtree sum
  int left = SBTUtil(root.left);
 
  // Find the right subtree sum
  int right = SBTUtil(root.right);
 
  // Condition of specially
  // balanced node
  if (root.left != null &&
      root.right != null)
  {
    // Condition of specially
    // balanced node
    if ((left % 2 == 0 && right % 2 != 0) ||
        (left % 2 != 0 && right % 2 == 0))
    {
      sum += root.data;
    }
  }
 
  // Return the sum
  return left + right + root.data;
}
 
//Function to build the binary tree
static Node build_tree(int R, int N,
                       String []str,
                       int []values)
{
  // Form root node of the tree
  Node root = newnode(R);
  int i;
 
  // Insert nodes into tree
  for (i = 0; i < N - 1; i++)
  {
    String s = str[i];
    int x = values[i];
 
    // Create a new Node
    Node temp = newnode(x);
 
    // Insert the node
    root = insert(s, 0, s.Length,
                  root, temp);
  }
 
  // Return the root of the Tree
  return root;
}
 
// Function to find the
// sum of specially
// balanced nodes
static void speciallyBalancedNodes(int R, int N,
                                   String []str,
                                   int []values)
{
  // Build Tree
  Node root = build_tree(R, N, str,
                         values);
 
  // Stores the sum of specially
  // balanced node
  sum = 0;
 
  // Function Call
  SBTUtil(root);
 
  // Print required sum
  Console.Write(sum + " ");
}
 
//Driver Code
public static void Main(String[] args)
{
  // Given nodes
  int N = 7;
 
  // Given root
  int R = 12;
 
  // Given path info of nodes
  // from root
  String []str = {"L", "R", "RL",
                  "RR", "RLL", "RLR"};
 
  // Given node values
  int []values = {17, 16, 4,
                  9, 2, 3};
 
  // Function Call
  speciallyBalancedNodes(R, N, str,
                         values);
}
}
 
// This code is contributed by 29AjayKumar

输出：

16

时间复杂度： O(N)
辅助空间： O(N)