# Sum of all the grandchild nodes of binary tree in kotlin

Kotlin program for Sum of all the grandchild nodes of binary tree. Here mentioned other language solution.

``````/*
Kotlin program for
Sum of nodes in binary tree whose grandparents exists
*/
// 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;
}
}
class BinaryTree
{
var root: TreeNode ? ;
constructor()
{
this.root = null;
}
fun preOrder(node: TreeNode ? ): Unit
{
if (node != null)
{
// Print node value
print("  " + node.data);
// Visit left subtree
this.preOrder(node.left);
// Visit right subtree
this.preOrder(node.right);
}
}
fun grandchildSum(node: TreeNode ? , level : Int): Int
{
if (node != null)
{
var result: Int = 0;
if (level > 2)
{
// Get the node
result = node.data;
}
// Visit left subtree and right subtree
return result +
this.grandchildSum(node.left, level + 1) +
this.grandchildSum(node.right, level + 1);
}
else
{
return 0;
}
}
}
fun main(args: Array < String > ): Unit
{
// Create new binary trees
val tree: BinaryTree = BinaryTree();
/*
4
/  \
/    \
12     7
/ \      \
2   3      1
/ \    /
6   8  5
/
9
----------------
Constructing binary tree
*/
tree.root = TreeNode(4);
tree.root?.left = TreeNode(12);
tree.root?.left?.right = TreeNode(3);
tree.root?.left?.right?.left = TreeNode(6);
tree.root?.left?.right?.left?.left = TreeNode(9);
tree.root?.left?.right?.right = TreeNode(8);
tree.root?.left?.left = TreeNode(2);
tree.root?.right = TreeNode(7);
tree.root?.right?.right = TreeNode(1);
tree.root?.right?.right?.left = TreeNode(5);
// Display tree node
print("\n Tree Nodes ");
tree.preOrder(tree.root);
// Find node sum
// 2 + 3 + 6 + 9 + 8 + 1 + 5 = 34
val result: Int = tree.grandchildSum(tree.root, 1);
// Display result
println("\n Result : " + result);
}``````

Output

`````` Tree Nodes   4  12  2  3  6  9  8  7  1  5
Result : 34``````

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