import commonclasses.TreeNode

/**
 * Given a binary search tree (BST), find the lowest common ancestor (LCA) node of two given nodes in the BST.
 *
 * According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”
 *
 *
 *
 * Example 1:
 *
 *
 * Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8
 * Output: 6
 * Explanation: The LCA of nodes 2 and 8 is 6.
 * Example 2:
 *
 *
 * Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4
 * Output: 2
 * Explanation: The LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.
 * Example 3:
 *
 * Input: root = [2,1], p = 2, q = 1
 * Output: 2
 *
 *
 * Constraints:
 *
 * The number of nodes in the tree is in the range [2, 10^5].
 * -10^9 <= Node.val <= 10^9
 * All Node.val are unique.
 * p != q
 * p and q will exist in the BST.
 * @see <a href="https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/">LeetCode</a>
 *
 * Definition for a binary tree node.
 * class TreeNode(var `val`: Int = 0) {
 *     var left: TreeNode? = null
 *     var right: TreeNode? = null
 * }
 */
fun lowestCommonAncestor(root: TreeNode?, p: TreeNode?, q: TreeNode?): TreeNode? {
    return when {
        root!!.`val` > p!!.`val` && root.`val` > q!!.`val` -> lowestCommonAncestor(root.left, p, q)
        root.`val` < p.`val` && root.`val` < q!!.`val` -> lowestCommonAncestor(root.right, p, q)
        else -> root
    }
}
