783. Minimum Distance Between BST Nodes

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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:

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

根据题目可知，这是一颗BST树，BST树的特点就是中序遍历是有序的，题目要求任意两点只差的最小值，其实就是看BST中序遍历后相邻节点只差的绝对值的最小值。

程序如下所示：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    long min = Integer.MAX_VALUE, pre = Integer.MIN_VALUE;
    public int minDiffInBST(TreeNode root) {
        inOrder(root);
        return (int)min;
    }
    public void inOrder(TreeNode root){
        if (root != null){
            inOrder(root.left);
            min = Math.min(min, Math.abs(root.val - pre));
            pre = root.val;
            inOrder(root.right);
        }
    }
}