LeetCode in Kotlin
98. Validate Binary Search Tree
Medium
Given the root of a binary tree, determine if it is a valid binary search tree (BST).
A valid 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:
Input: root = [2,1,3]
Output: true
Example 2:
Input: root = [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.
Constraints:
- The number of nodes in the tree is in the range
[1, 104]. -231 <= Node.val <= 231 - 1
Solution
import com_github_leetcode.TreeNode
/*
* 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 {
return solve(root, Long.MIN_VALUE, Long.MAX_VALUE)
}
// we will send a valid range and check whether the root lies in the range
// and update the range for the subtrees
private fun solve(root: TreeNode?, left: Long, right: Long): Boolean {
if (root == null) {
return true
}
return if (root.`val` <= left || root.`val` >= right) {
false
} else {
solve(root.left, left, root.`val`.toLong()) && solve(root.right, root.`val`.toLong(), right)
}
}
}