LeetCode in Kotlin
783. Minimum Distance Between BST Nodes
Easy
Given the root of a Binary Search Tree (BST), return the minimum difference between the values of any two different nodes in the tree.
Example 1:
Input: root = [4,2,6,1,3]
Output: 1
Example 2:
Input: root = [1,0,48,null,null,12,49]
Output: 1
Constraints:
- The number of nodes in the tree is in the range
[2, 100]. 0 <= Node.val <= 105
Note: This question is the same as 530: https://leetcode.com/problems/minimum-absolute-difference-in-bst/
Solution
import com_github_leetcode.TreeNode
import kotlin.math.abs
/*
* 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 {
private var prev = -1
private var min = Int.MAX_VALUE
fun minDiffInBST(root: TreeNode?): Int {
if (root == null) {
return min
}
minDiffInBST(root.left)
if (prev != -1) {
min = min.coerceAtMost(abs(root.`val` - prev))
}
prev = root.`val`
minDiffInBST(root.right)
return min
}
}