98.validate-binary-search-tree.kt

/*
 * @lc app=leetcode id=98 lang=kotlin
 *
 * [98] Validate Binary Search Tree
 *
 * https://leetcode.com/problems/validate-binary-search-tree/description/
 *
 * algorithms
 * Medium (26.08%)
 * Likes:    2177
 * Dislikes: 327
 * Total Accepted:    442.1K
 * Total Submissions: 1.7M
 * Testcase Example:  '[2,1,3]'
 *
 * Given a binary tree, determine if it is a valid binary search tree (BST).
 * 
 * Assume a BST is defined as follows:
 * 
 * 
 * The left subtree of a node contains only nodes with keys less than the
 * node's key.
 * The right subtree of a node contains only nodes with keys greater than the
 * node's key.
 * Both the left and right subtrees must also be binary search trees.
 * 
 * 
 * 
 * 
 * Example 1:
 * 
 * 
 * ⁠   2
 * ⁠  / \
 * ⁠ 1   3
 * 
 * Input: [2,1,3]
 * Output: true
 * 
 * 
 * Example 2:
 * 
 * 
 * ⁠   5
 * ⁠  / \
 * ⁠ 1   4
 * / \
 * 3   6
 * 
 * Input: [5,1,4,null,null,3,6]
 * Output: false
 * Explanation: The root node's value is 5 but its right child's value is 4.
 * 
 * 
 */
/**
 * Example:
 * var ti = TreeNode(5)
 * var v = ti.`val`
 * Definition for a binary tree node.
 * class TreeNode(var `val`: Int) {
 *     var left: TreeNode? = null
 *     var right: TreeNode? = null
 * }
 */
class Solution {
    fun isValidBST(root: TreeNode?): Boolean {
        if (root == null) return true
        return valid(root, Long.MIN_VALUE, Long.MAX_VALUE)
    }

    private fun valid(root: TreeNode?, low: Long, high: Long): Boolean {
        if (root == null) return true
        if (root!!.`val` <= low || root!!.`val` >= high) return false
        return valid(root?.left, low, root!!.`val`.toLong()) && valid(root?.right, root!!.`val`.toLong(), high);
    }
}