Sum of specially balanced nodes from a given Binary Tree

Given a Binary Tree, the task is to find the sum of all the specially balanced nodes in the given Binary Tree.

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.

Examples:

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.

Approach: The idea is to perform DFS Traversal on the given Tree using recursion and update the final sum as per the given conditions. Follow the steps below:

  1. Initialize a totalSum as 0 which stores the sum of all specially balanced nodes.
  2. Perform the DFS Traversal on the given Tree and check for the following:
    • If the node is a leaf node, then return the value of that node.
    • Now, if the current node is not a leaf node, recursively traverse for the left and right subtree.
    • Return the value of the sum of the left and right subtree with the current root value when each recursive call ends.
    • Check if the returned sums satisfy the property of the specially balanced node. IF found to be true, add the current node value to totalSum.
  3. Print the value of totalSum after completing the above steps.

Below is the implementation of the above approach:

C++

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// C++ program for the above approach
#include <bits/stdc++.h>
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;
}

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Java

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// 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

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Python3

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# 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

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C#

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//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

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Output: 

16










 

Time Complexity: O(N)
Auxiliary Space: O(N)

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