Check for continuous binary tree in kotlin
Kotlin program for Check for continuous binary tree. Here problem description and explanation.
/*
Kotlin program for
Check if binary tree is continuous tree
Using recursion
*/
// 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 absValue(num: Int): Int
{
if (num < 0)
{
return -num;
}
return num;
}
// Check tree is continuous or not
fun isContinuous(node: TreeNode ? ): Boolean
{
if (node != null)
{
if ((node.left != null &&
this.absValue(node.data - node.left!!.data) != 1) ||
(node.right != null &&
this.absValue(node.data - node.right!!.data) != 1))
{
// Case
// When fail continuous tree rule
return false;
}
if (this.isContinuous(node.left) &&
this.isContinuous(node.right))
{
return true;
}
// When node value is not satisfied
// continuous tree properties
return false;
}
return true;
}
}
fun main(args: Array < String > ): Unit
{
// Create new tree
val tree: BinaryTree = BinaryTree();
/*
Binary Tree
------------
5
/ \
4 4
/ / \
3 5 3
\
2
*/
// Add tree node
tree.root = TreeNode(5);
tree.root?.left = TreeNode(4);
tree.root?.right = TreeNode(4);
tree.root?.left?.left = TreeNode(3);
tree.root?.right?.right = TreeNode(3);
tree.root?.right?.left = TreeNode(5);
tree.root?.left?.left?.right = TreeNode(2);
if (tree.isContinuous(tree.root) == true)
{
println("Continuous Tree ");
}
else
{
println("Not Continuous Tree ");
}
// Case 2
/*
5
/ \
4 4
/ / \
3 5 3
\
1 <--- change value
*/
tree.root?.left?.left?.right?.data = 1;
if (tree.isContinuous(tree.root) == true)
{
println("Continuous Tree ");
}
else
{
println("Not Continuous Tree ");
}
}
Output
Continuous Tree
Not Continuous Tree
Please share your knowledge to improve code and content standard. Also submit your doubts, and test case. We improve by your feedback. We will try to resolve your query as soon as possible.
New Comment