how can I translate this to Kotlin? ```class Solu...

# getting-started

[oday]

how can I translate this to Kotlin?

class Solution(object):
  def _isValidBSTHelper(self, n, low, high):
    if not n:
      return True
    val = n.val
    if ((val > low and val < high) and
        self._isValidBSTHelper(n.left, low, n.val) and
        self._isValidBSTHelper(n.right, n.val, high)):
        return True
    return False

  def isValidBST(self, n):
    return self._isValidBSTHelper(n, float('-inf'), float('inf'))

my attempt produced this

fun isValidChecker(
    node: Node<Int>?, low: Int, high: Int
): Boolean {
    if (node == null)
        return true

    val value = node.value

    if (value in (low + 1) until high
        && isValidChecker(node.left, low, value)
        && isValidChecker(node.right, value, high)
    )
        return true

    return false
}

fun isValidBST(node: Node<Int>?): Boolean {
    return isValidChecker(
        node, Int.MIN_VALUE, Int.MAX_VALUE
    )
}

[oday]

issue is what to feed isValidBST

[oday]

I want negative infinity and positive infinity with the current way that the code looks

[oday]

I’m with you logically, but running the above Python variation of it makes this LeetCode question pass, running our Kotlin version fails on 7 cases

[oday]

so something is off

[oday]

https://leetcode.com/problems/validate-binary-search-tree/

[oday]

can’t we use an operator that is exclusive?

[oday]

instead of the +1 and -1 ?

[oday]

dont get that part

[oday]

think I broke it

class Solution {
    fun TreeNode?.isValidBST(
        low: Double = Double.NEGATIVE_INFINITY, high: Double = Double.POSITIVE_INFINITY
    ): Boolean = this == null || 
    (`val`.toDouble() > low && `val`.toDouble() < high)
    && left.isValidBST(low, `val`.toDouble())
    && right.isValidBST(`val`.toDouble(), high)

    fun isValidBST(root: TreeNode?): Boolean {
        return root.isValidBST()
    }
}

[oday]

this works on Leetcode

[oday]

i insisted on having it as an extension lol