LeetCode之Minimum Distance Between BST Nodes（Kotlin）

問題： Given a Binary Search Tree (BST) with the root node root, return the minimum difference between the values of any two different nodes in the tree.

Example :

Input: root = [4,2,6,1,3,null,null]
Output: 1
Explanation:
Note that root is a TreeNode object, not an array.

The given tree [4,2,6,1,3,null,null] is represented by the following diagram:

          4
        /   \
      2      6
     / \    
    1   3  

while the minimum difference in this tree is 1, it occurs between node 1 and node 2, also between node 3 and node 2.
Note:

The size of the BST will be between 2 and 100.
The BST is always valid, each node's value is an integer, and each node's value is different.
複製代碼

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方法： 題目中的二叉搜索樹節點間具備特殊的大小關係，即左節點<根節點<右節點，則經過中序遍歷就是從小到大排序的節點，而後計算節點間的diff，取最小diff即爲最終結果。

具體實現：

class MinimumDistanceBetweenBSTNodes {

    // Definition for a binary tree node.
    class TreeNode(var `val`: Int) {
        var left: TreeNode? = null
        var right: TreeNode? = null
    }

    private var pre = -100
    private var diff = 100

    fun minDiffInBST(root: TreeNode?): Int {
        if (root == null) {
            return Int.MIN_VALUE
        }
        minDiffInBST(root.left)
        diff = minOf(diff, root.`val` - pre)
        pre = root.`val`
        minDiffInBST(root.right)
        return diff
    }
}

fun main(args: Array<String>) {
    val tree4 = MinimumDistanceBetweenBSTNodes.TreeNode(4)
    val tree2 = MinimumDistanceBetweenBSTNodes.TreeNode(2)
    val tree6 = MinimumDistanceBetweenBSTNodes.TreeNode(6)
    val tree1 = MinimumDistanceBetweenBSTNodes.TreeNode(1)
    val tree3 = MinimumDistanceBetweenBSTNodes.TreeNode(3)
    tree4.left = tree2
    tree4.right = tree6
    tree2.left = tree1
    tree2.right = tree3
    val minimumDistanceBetweenBSTNodes = MinimumDistanceBetweenBSTNodes()
    println(minimumDistanceBetweenBSTNodes.minDiffInBST(tree4))
}
複製代碼

有問題隨時溝通

具體代碼實現能夠參考Github