# Sum of nodes having sum of subtrees of opposition parties

Here given code implementation process.

``````// C program for
// Sum of nodes having sum of subtrees of opposition parties
#include <stdio.h>

#include <stdlib.h>

// Tree Node
struct TreeNode
{
int data;
struct TreeNode *left;
struct TreeNode *right;
};
// Binary Tree
struct BinaryTree
{
struct TreeNode *root;
};
// Create new tree
struct BinaryTree *new_tree()
{
// Create dynamic node
struct BinaryTree *tree = (struct BinaryTree *) malloc(sizeof(struct BinaryTree));
if (tree != NULL)
{
tree->root = NULL;
}
else
{
printf("Memory Overflow to Create tree Tree\n");
}
//return new tree
return tree;
}
// This is creates and returns the new binary tree node
struct TreeNode *get_node(int data)
{
// Create dynamic node
struct TreeNode *new_node = (struct TreeNode *) malloc(sizeof(struct TreeNode));
if (new_node != NULL)
{
//Set data and pointer values
new_node->data = data;
new_node->left = NULL;
new_node->right = NULL;
}
else
{
//This is indicates, segmentation fault or memory overflow problem
printf("Memory Overflow\n");
}
//return new node
return new_node;
}
// Calculate sum of all node which are left and right
// subtree have Even and Odd sum
int node_sum(struct TreeNode *node, int *sum)
{
if (node == NULL)
{
return 0;
}
// Find left and right subtree sum
int l = node_sum(node->left, sum);
int r = node_sum(node->right, sum);
if (node->left != NULL && node->right != NULL && (l + r) % 2 != 0)
{
// When node have both exist left and right subtree
// and its left and right subtree sum are even odd
*sum = *sum + node->data;
}
// returns the subtree sum
return l + r + node->data;
}
// Handles the request to find sum of opposite subtree
void opposite_sum(struct TreeNode *root)
{
if (root == NULL)
{
printf(" Empty Tree\n");
}
else
{
int sum = 0;
node_sum(root, & sum);
// Display calculate sum
printf(" Sum : %d\n", sum);
}
}
int main(int argc, char const *argv[])
{
struct BinaryTree *tree = new_tree();
/*
constructor binary tree
-----------------
6
/   \
4     7
/ \     \
2   3     12
/ \
10  8
\
-1

-----------------
First Tree

*/
tree->root = get_node(6);
tree->root->left = get_node(4);
tree->root->left->right = get_node(3);
tree->root->left->right->left = get_node(10);
tree->root->left->right->right = get_node(8);
tree->root->left->right->right->right = get_node(-1);
tree->root->left->left = get_node(2);
tree->root->right = get_node(7);
tree->root->right->right = get_node(12);
/*

Adding the left and right subtree value
-----------------
45
/   \
26    19
/ \     \
2  20     12
/ \
10  7
\
-1
-----------------
Second tree

Resultant nodes in First tree
—————————————————————————————
[3] Because left subtree sum 10 and right subtree sum have 7
[6] Because left subtree sum 26 and right subtree sum have 19
—————
[9]  Output

*/
opposite_sum(tree->root);
return 0;
}``````

#### input

`` Sum : 9``
``````/*
Java Program for
Sum of nodes having sum of subtrees of opposition parties
*/
// Binary Tree node
class TreeNode
{
public int data;
public TreeNode left;
public TreeNode right;
public TreeNode(int data)
{
// Set node value
this.data = data;
this.left = null;
this.right = null;
}
}
// Define Binary Tree
public class BinaryTree
{
public TreeNode root;
public int sum;
public BinaryTree()
{
this.root = null;
this.sum = 0;
}
// Calculate sum of all node which are left and right
// subtree have Even and Odd sum
public int nodeSum(TreeNode node)
{
if (node == null)
{
return 0;
}
// Find left and right subtree sum
int l = nodeSum(node.left);
int r = nodeSum(node.right);
if (node.left != null && node.right != null && (l + r) % 2 != 0)
{
// When node have both exist left and right subtree
// and its left and right subtree sum are even odd
this.sum = this.sum + node.data;
}
// returns the subtree sum
return l + r + node.data;
}
// Handles the request to find sum of opposite subtree
public void oppositeSum()
{
if (this.root == null)
{
System.out.println(" Empty Tree");
}
else
{
this.sum = 0;
this.nodeSum(this.root);
// Display calculate sum
System.out.println(" Sum : " + sum);
}
}
public static void main(String[] args)
{
BinaryTree tree = new BinaryTree();
/*
constructor binary tree
-----------------
6
/   \
4     7
/ \     \
2   3     12
/ \
10  8
\
-1

-----------------
First Tree

*/
tree.root = new TreeNode(6);
tree.root.left = new TreeNode(4);
tree.root.left.right = new TreeNode(3);
tree.root.left.right.left = new TreeNode(10);
tree.root.left.right.right = new TreeNode(8);
tree.root.left.right.right.right = new TreeNode(-1);
tree.root.left.left = new TreeNode(2);
tree.root.right = new TreeNode(7);
tree.root.right.right = new TreeNode(12);
/*

Adding the left and right subtree value
-----------------
45
/   \
26    19
/ \     \
2  20     12
/ \
10  7
\
-1
-----------------
Second tree

Resultant nodes in First tree
—————————————————————————————
[3] Because left subtree sum 10 and right subtree sum have 7
[6] Because left subtree sum 26 and right subtree sum have 19
—————
[9]  Output

*/
tree.oppositeSum();
}
}``````

#### input

`` Sum : 9``
``````// Include header file
#include <iostream>
using namespace std;
/*
C++ Program for
Sum of nodes having sum of subtrees of opposition parties
*/
// Binary Tree node
class TreeNode
{
public:
int data;
TreeNode *left;
TreeNode *right;
TreeNode(int data)
{
// Set node value
this->data = data;
this->left = NULL;
this->right = NULL;
}
};
// Define Binary Tree
class BinaryTree
{
public:
TreeNode *root;
int sum;
BinaryTree()
{
this->root = NULL;
this->sum = 0;
}
// Calculate sum of all node which are left and right
// subtree have Even and Odd sum
int nodeSum(TreeNode *node)
{
if (node == NULL)
{
return 0;
}
// Find left and right subtree sum
int l = this->nodeSum(node->left);
int r = this->nodeSum(node->right);
if (node->left != NULL && node->right != NULL && (l + r) % 2 != 0)
{
// When node have both exist left and right subtree
// and its left and right subtree sum are even odd
this->sum = this->sum + node->data;
}
// returns the subtree sum
return l + r + node->data;
}
// Handles the request to find sum of opposite subtree
void oppositeSum()
{
if (this->root == NULL)
{
cout << " Empty Tree" << endl;
}
else
{
this->sum = 0;
this->nodeSum(this->root);
// Display calculate sum
cout << " Sum : " << this->sum << endl;
}
}
};
int main()
{
BinaryTree *tree = new BinaryTree();
/*
constructor binary tree
-----------------
6
/   \
4     7
/ \     \
2   3     12
/ \
10  8
\
-1
-----------------
First Tree
*/
tree->root = new TreeNode(6);
tree->root->left = new TreeNode(4);
tree->root->left->right = new TreeNode(3);
tree->root->left->right->left = new TreeNode(10);
tree->root->left->right->right = new TreeNode(8);
tree->root->left->right->right->right = new TreeNode(-1);
tree->root->left->left = new TreeNode(2);
tree->root->right = new TreeNode(7);
tree->root->right->right = new TreeNode(12);
/*
Adding the left and right subtree value
-----------------
45
/   \
26    19
/ \     \
2  20     12
/ \
10  7
\
-1
-----------------
Second tree

Resultant nodes in First tree
—————————————————————————————
[3] Because left subtree sum 10 and right subtree sum have 7
[6] Because left subtree sum 26 and right subtree sum have 19
—————
[9]  Output
*/
tree->oppositeSum();
return 0;
}``````

#### input

`` Sum : 9``
``````// Include namespace system
using System;
/*
Csharp Program for
Sum of nodes having sum of subtrees of opposition parties
*/
// Binary Tree node
public class TreeNode
{
public int data;
public TreeNode left;
public TreeNode right;
public TreeNode(int data)
{
// Set node value
this.data = data;
this.left = null;
this.right = null;
}
}
// Define Binary Tree
public class BinaryTree
{
public TreeNode root;
public int sum;
public BinaryTree()
{
this.root = null;
this.sum = 0;
}
// Calculate sum of all node which are left and right
// subtree have Even and Odd sum
public int nodeSum(TreeNode node)
{
if (node == null)
{
return 0;
}
// Find left and right subtree sum
int l = this.nodeSum(node.left);
int r = this.nodeSum(node.right);
if (node.left != null && node.right != null && (l + r) % 2 != 0)
{
// When node have both exist left and right subtree
// and its left and right subtree sum are even odd
this.sum = this.sum + node.data;
}
// returns the subtree sum
return l + r + node.data;
}
// Handles the request to find sum of opposite subtree
public void oppositeSum()
{
if (this.root == null)
{
Console.WriteLine(" Empty Tree");
}
else
{
this.sum = 0;
this.nodeSum(this.root);
// Display calculate sum
Console.WriteLine(" Sum : " + this.sum);
}
}
public static void Main(String[] args)
{
BinaryTree tree = new BinaryTree();
/*
constructor binary tree
-----------------
6
/   \
4     7
/ \     \
2   3     12
/ \
10  8
\
-1
-----------------
First Tree
*/
tree.root = new TreeNode(6);
tree.root.left = new TreeNode(4);
tree.root.left.right = new TreeNode(3);
tree.root.left.right.left = new TreeNode(10);
tree.root.left.right.right = new TreeNode(8);
tree.root.left.right.right.right = new TreeNode(-1);
tree.root.left.left = new TreeNode(2);
tree.root.right = new TreeNode(7);
tree.root.right.right = new TreeNode(12);
/*
Adding the left and right subtree value
-----------------
45
/   \
26    19
/ \     \
2  20     12
/ \
10  7
\
-1
-----------------
Second tree

Resultant nodes in First tree
—————————————————————————————
[3] Because left subtree sum 10 and right subtree sum have 7
[6] Because left subtree sum 26 and right subtree sum have 19
—————
[9]  Output
*/
tree.oppositeSum();
}
}``````

#### input

`` Sum : 9``
``````<?php
/*
Php Program for
Sum of nodes having sum of subtrees of opposition parties
*/
// Binary Tree node
class TreeNode
{
public \$data;
public \$left;
public \$right;
public	function __construct(\$data)
{
// Set node value
\$this->data = \$data;
\$this->left = NULL;
\$this->right = NULL;
}
}
// Define Binary Tree
class BinaryTree
{
public \$root;
public \$sum;
public	function __construct()
{
\$this->root = NULL;
\$this->sum = 0;
}
// Calculate sum of all node which are left and right
// subtree have Even and Odd sum
public	function nodeSum(\$node)
{
if (\$node == NULL)
{
return 0;
}
// Find left and right subtree sum
\$l = \$this->nodeSum(\$node->left);
\$r = \$this->nodeSum(\$node->right);
if (\$node->left != NULL && \$node->right != NULL && (\$l + \$r) % 2 != 0)
{
// When node have both exist left and right subtree
// and its left and right subtree sum are even odd
\$this->sum = \$this->sum + \$node->data;
}
// returns the subtree sum
return \$l + \$r + \$node->data;
}
// Handles the request to find sum of opposite subtree
public	function oppositeSum()
{
if (\$this->root == NULL)
{
echo " Empty Tree\n";
}
else
{
\$this->sum = 0;
\$this->nodeSum(\$this->root);
// Display calculate sum
echo " Sum : ".\$this->sum."\n";
}
}
}

function main()
{
\$tree = new BinaryTree();
/*
constructor binary tree
-----------------
6
/   \
4     7
/ \     \
2   3     12
/ \
10  8
\
-1
-----------------
First Tree
*/
\$tree->root = new TreeNode(6);
\$tree->root->left = new TreeNode(4);
\$tree->root->left->right = new TreeNode(3);
\$tree->root->left->right->left = new TreeNode(10);
\$tree->root->left->right->right = new TreeNode(8);
\$tree->root->left->right->right->right = new TreeNode(-1);
\$tree->root->left->left = new TreeNode(2);
\$tree->root->right = new TreeNode(7);
\$tree->root->right->right = new TreeNode(12);
/*
Adding the left and right subtree value
-----------------
45
/   \
26    19
/ \     \
2  20     12
/ \
10  7
\
-1
-----------------
Second tree

Resultant nodes in First tree
—————————————————————————————
[3] Because left subtree sum 10 and right subtree sum have 7
[6] Because left subtree sum 26 and right subtree sum have 19
—————
[9]  Output
*/
\$tree->oppositeSum();
}
main();``````

#### input

`` Sum : 9``
``````/*
Node JS Program for
Sum of nodes having sum of subtrees of opposition parties
*/
// Binary Tree node
class TreeNode
{
constructor(data)
{
// Set node value
this.data = data;
this.left = null;
this.right = null;
}
}
// Define Binary Tree
class BinaryTree
{
constructor()
{
this.root = null;
this.sum = 0;
}
// Calculate sum of all node which are left and right
// subtree have Even and Odd sum
nodeSum(node)
{
if (node == null)
{
return 0;
}
// Find left and right subtree sum
var l = this.nodeSum(node.left);
var r = this.nodeSum(node.right);
if (node.left != null && node.right != null && (l + r) % 2 != 0)
{
// When node have both exist left and right subtree
// and its left and right subtree sum are even odd
this.sum = this.sum + node.data;
}
// returns the subtree sum
return l + r + node.data;
}
// Handles the request to find sum of opposite subtree
oppositeSum()
{
if (this.root == null)
{
console.log(" Empty Tree");
}
else
{
this.sum = 0;
this.nodeSum(this.root);
// Display calculate sum
console.log(" Sum : " + this.sum);
}
}
}

function main()
{
var tree = new BinaryTree();
/*
constructor binary tree
-----------------
6
/   \
4     7
/ \     \
2   3     12
/ \
10  8
\
-1
-----------------
First Tree
*/
tree.root = new TreeNode(6);
tree.root.left = new TreeNode(4);
tree.root.left.right = new TreeNode(3);
tree.root.left.right.left = new TreeNode(10);
tree.root.left.right.right = new TreeNode(8);
tree.root.left.right.right.right = new TreeNode(-1);
tree.root.left.left = new TreeNode(2);
tree.root.right = new TreeNode(7);
tree.root.right.right = new TreeNode(12);
/*
Adding the left and right subtree value
-----------------
45
/   \
26    19
/ \     \
2  20     12
/ \
10  7
\
-1
-----------------
Second tree

Resultant nodes in First tree
—————————————————————————————
[3] Because left subtree sum 10 and right subtree sum have 7
[6] Because left subtree sum 26 and right subtree sum have 19
—————
[9]  Output
*/
tree.oppositeSum();
}
main();``````

#### input

`` Sum : 9``
``````#  Python 3 Program for
#  Sum of nodes having sum of subtrees of opposition parties

#  Binary Tree node
class TreeNode :
def __init__(self, data) :
#  Set node value
self.data = data
self.left = None
self.right = None

#  Define Binary Tree
class BinaryTree :
def __init__(self) :
self.root = None
self.sum = 0

#  Calculate sum of all node which are left and right
#  subtree have Even and Odd sum
def nodeSum(self, node) :
if (node == None) :
return 0

l = self.nodeSum(node.left)
r = self.nodeSum(node.right)
if (node.left != None and node.right != None and(l + r) % 2 != 0) :
#  When node have both exist left and right subtree
#  and its left and right subtree sum are even odd
self.sum = self.sum + node.data

#  returns the subtree sum
return l + r + node.data

#  Handles the request to find sum of opposite subtree
def oppositeSum(self) :
if (self.root == None) :
print(" Empty Tree")
else :
self.sum = 0
self.nodeSum(self.root)
#  Display calculate sum
print(" Sum : ", self.sum)

def main() :
tree = BinaryTree()
#    constructor binary tree
#    -----------------
#         6
#       /   \
#      4     7
#     / \     \
#    2   3     12
#       / \
#      10  8
#           \
#           -1
#    -----------------
#    First Tree
tree.root = TreeNode(6)
tree.root.left = TreeNode(4)
tree.root.left.right = TreeNode(3)
tree.root.left.right.left = TreeNode(10)
tree.root.left.right.right = TreeNode(8)
tree.root.left.right.right.right = TreeNode(-1)
tree.root.left.left = TreeNode(2)
tree.root.right = TreeNode(7)
tree.root.right.right = TreeNode(12)
#    Adding the left and right subtree value
#    -----------------
#         45
#       /   \
#      26    19
#     / \     \
#    2  20     12
#       / \
#      10  7
#           \
#           -1
#    -----------------
#    Second tree
#    Resultant nodes in First tree
#    —————————————————————————————
#    [3] Because left subtree sum 10 and right subtree sum have 7
#    [6] Because left subtree sum 26 and right subtree sum have 19
#    —————
#    [9]  Output
tree.oppositeSum()

if __name__ == "__main__": main()``````

#### input

`` Sum :  9``
``````#  Ruby Program for
#  Sum of nodes having sum of subtrees of opposition parties

#  Binary Tree node
class TreeNode
# Define the accessor and reader of class TreeNode
attr_accessor :data, :left, :right
def initialize(data)
#  Set node value
self.data = data
self.left = nil
self.right = nil
end

end

#  Define Binary Tree
class BinaryTree
# Define the accessor and reader of class BinaryTree
attr_accessor :root, :sum
def initialize()
self.root = nil
self.sum = 0
end

#  Calculate sum of all node which are left and right
#  subtree have Even and Odd sum
def nodeSum(node)
if (node == nil)
return 0
end

#  Find left and right subtree sum
l = self.nodeSum(node.left)
r = self.nodeSum(node.right)
if (node.left != nil && node.right != nil && (l + r) % 2 != 0)
#  When node have both exist left and right subtree
#  and its left and right subtree sum are even odd
self.sum = self.sum + node.data
end

#  returns the subtree sum
return l + r + node.data
end

#  Handles the request to find sum of opposite subtree
def oppositeSum()
if (self.root == nil)
print(" Empty Tree", "\n")
else
self.sum = 0
self.nodeSum(self.root)
#  Display calculate sum
print(" Sum : ", self.sum, "\n")
end

end

end

def main()
tree = BinaryTree.new()
#    constructor binary tree
#    -----------------
#         6
#       /   \
#      4     7
#     / \     \
#    2   3     12
#       / \
#      10  8
#           \
#           -1
#    -----------------
#    First Tree
tree.root = TreeNode.new(6)
tree.root.left = TreeNode.new(4)
tree.root.left.right = TreeNode.new(3)
tree.root.left.right.left = TreeNode.new(10)
tree.root.left.right.right = TreeNode.new(8)
tree.root.left.right.right.right = TreeNode.new(-1)
tree.root.left.left = TreeNode.new(2)
tree.root.right = TreeNode.new(7)
tree.root.right.right = TreeNode.new(12)
#    Adding the left and right subtree value
#    -----------------
#         45
#       /   \
#      26    19
#     / \     \
#    2  20     12
#       / \
#      10  7
#           \
#           -1
#    -----------------
#    Second tree
#    Resultant nodes in First tree
#    —————————————————————————————
#    [3] Because left subtree sum 10 and right subtree sum have 7
#    [6] Because left subtree sum 26 and right subtree sum have 19
#    —————
#    [9]  Output
tree.oppositeSum()
end

main()``````

#### input

`````` Sum : 9
``````
``````/*
Scala Program for
Sum of nodes having sum of subtrees of opposition parties
*/
// Binary Tree node
class TreeNode(var data: Int , var left: TreeNode , var right: TreeNode)
{
def this(data: Int)
{
// Set node value
this(data,null,null);
}
}
// Define Binary Tree
class BinaryTree(var root: TreeNode , var sum: Int)
{
def this()
{
this(null, 0);
}
// Calculate sum of all node which are left and right
// subtree have Even and Odd sum
def nodeSum(node: TreeNode): Int = {
if (node == null)
{
return 0;
}
// Find left and right subtree sum
var l: Int = nodeSum(node.left);
var r: Int = nodeSum(node.right);
if (node.left != null && node.right != null && (l + r) % 2 != 0)
{
// When node have both exist left and right subtree
// and its left and right subtree sum are even odd
this.sum = this.sum + node.data;
}
// returns the subtree sum
return l + r + node.data;
}
// Handles the request to find sum of opposite subtree
def oppositeSum(): Unit = {
if (this.root == null)
{
println(" Empty Tree");
}
else
{
this.sum = 0;
this.nodeSum(this.root);
// Display calculate sum
println(" Sum : " + sum);
}
}
}
object Main
{
def main(args: Array[String]): Unit = {
var tree: BinaryTree = new BinaryTree();
/*
constructor binary tree
-----------------
6
/   \
4     7
/ \     \
2   3     12
/ \
10  8
\
-1
-----------------
First Tree
*/
tree.root = new TreeNode(6);
tree.root.left = new TreeNode(4);
tree.root.left.right = new TreeNode(3);
tree.root.left.right.left = new TreeNode(10);
tree.root.left.right.right = new TreeNode(8);
tree.root.left.right.right.right = new TreeNode(-1);
tree.root.left.left = new TreeNode(2);
tree.root.right = new TreeNode(7);
tree.root.right.right = new TreeNode(12);
/*
Adding the left and right subtree value
-----------------
45
/   \
26    19
/ \     \
2  20     12
/ \
10  7
\
-1
-----------------
Second tree

Resultant nodes in First tree
—————————————————————————————
[3] Because left subtree sum 10 and right subtree sum have 7
[6] Because left subtree sum 26 and right subtree sum have 19
—————
[9]  Output
*/
tree.oppositeSum();
}
}``````

#### input

`` Sum : 9``
``````/*
Swift 4 Program for
Sum of nodes having sum of subtrees of opposition parties
*/
// Binary Tree node
class TreeNode
{
var data: Int;
var left: TreeNode? ;
var right: TreeNode? ;
init(_ data: Int)
{
// Set node value
self.data = data;
self.left = nil;
self.right = nil;
}
}
// Define Binary Tree
class BinaryTree
{
var root: TreeNode? ;
var sum: Int;
init()
{
self.root = nil;
self.sum = 0;
}
// Calculate sum of all node which are left and right
// subtree have Even and Odd sum
func nodeSum(_ node: TreeNode? )->Int
{
if (node == nil)
{
return 0;
}
// Find left and right subtree sum
let l: Int = self.nodeSum(node!.left);
let r: Int = self.nodeSum(node!.right);
if (node!.left  != nil && node!.right  != nil && (l + r) % 2  != 0)
{
// When node have both exist left and right subtree
// and its left and right subtree sum are even odd
self.sum = self.sum + node!.data;
}
// returns the subtree sum
return l + r + node!.data;
}
// Handles the request to find sum of opposite subtree
func oppositeSum()
{
if (self.root == nil)
{
print(" Empty Tree");
}
else
{
self.sum = 0;
let _ = self.nodeSum(self.root);
// Display calculate sum
print(" Sum : ", self.sum);
}
}
}
func main()
{
let tree: BinaryTree = BinaryTree();
/*
constructor binary tree
-----------------
6
/   \
4     7
/ \     \
2   3     12
/ \
10  8
\
-1
-----------------
First Tree
*/
tree.root = TreeNode(6);
tree.root!.left = TreeNode(4);
tree.root!.left!.right = TreeNode(3);
tree.root!.left!.right!.left = TreeNode(10);
tree.root!.left!.right!.right = TreeNode(8);
tree.root!.left!.right!.right!.right = TreeNode(-1);
tree.root!.left!.left = TreeNode(2);
tree.root!.right = TreeNode(7);
tree.root!.right!.right = TreeNode(12);
/*
Adding the left and right subtree value
-----------------
45
/   \
26    19
/ \     \
2  20     12
/ \
10  7
\
-1
-----------------
Second tree

Resultant nodes in First tree
—————————————————————————————
[3] Because left subtree sum 10 and right subtree sum have 7
[6] Because left subtree sum 26 and right subtree sum have 19
—————
[9]  Output
*/
tree.oppositeSum();
}
main();``````

#### input

`` Sum :  9``
``````/*
Kotlin Program for
Sum of nodes having sum of subtrees of opposition parties
*/
// Binary Tree node
class TreeNode
{
var data: Int;
var left: TreeNode ? ;
var right: TreeNode ? ;
constructor(data: Int)
{
// Set node value
this.data = data;
this.left = null;
this.right = null;
}
}
// Define Binary Tree
class BinaryTree
{
var root: TreeNode ? ;
var sum: Int;
constructor()
{
this.root = null;
this.sum = 0;
}
// Calculate sum of all node which are left and right
// subtree have Even and Odd sum
fun nodeSum(node: TreeNode ? ): Int
{
if (node == null)
{
return 0;
}
// Find left and right subtree sum
val l: Int = this.nodeSum(node.left);
val r: Int = this.nodeSum(node.right);
if (node.left != null && node.right != null && (l + r) % 2 != 0)
{
// When node have both exist left and right subtree
// and its left and right subtree sum are even odd
this.sum = this.sum + node.data;
}
// returns the subtree sum
return l + r + node.data;
}
// Handles the request to find sum of opposite subtree
fun oppositeSum(): Unit
{
if (this.root == null)
{
println(" Empty Tree");
}
else
{
this.sum = 0;
this.nodeSum(this.root);
// Display calculate sum
println(" Sum : " + this.sum);
}
}
}
fun main(args: Array < String > ): Unit
{
val tree: BinaryTree = BinaryTree();
/*
constructor binary tree
-----------------
6
/   \
4     7
/ \     \
2   3     12
/ \
10  8
\
-1
-----------------
First Tree
*/
tree.root = TreeNode(6);
tree.root?.left = TreeNode(4);
tree.root?.left?.right = TreeNode(3);
tree.root?.left?.right?.left = TreeNode(10);
tree.root?.left?.right?.right = TreeNode(8);
tree.root?.left?.right?.right?.right = TreeNode(-1);
tree.root?.left?.left = TreeNode(2);
tree.root?.right = TreeNode(7);
tree.root?.right?.right = TreeNode(12);
/*
Adding the left and right subtree value
-----------------
45
/   \
26    19
/ \     \
2  20     12
/ \
10  7
\
-1
-----------------
Second tree

Resultant nodes in First tree
—————————————————————————————
[3] Because left subtree sum 10 and right subtree sum have 7
[6] Because left subtree sum 26 and right subtree sum have 19
—————
[9]  Output
*/
tree.oppositeSum();
}``````

#### input

`` Sum : 9``

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