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题目：给你一个二叉搜索树的根节点 root ，返回 树中任意两不同节点值之间的最小差值 。

注意：本题与 530：https://leetcode-cn.com/problems/minimum-absolute-difference-in-bst/ 相同

示例 1：

输入：root = [4,2,6,1,3]

输入：root = [1,0,48,null,null,12,49]

提示：

树中节点数目在范围 [2, 100] 内

答案：

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode() {} *     TreeNode(int val) { this.val = val; } *     TreeNode(int val, TreeNode left, TreeNode right) { *         this.val = val; *         this.left = left; *         this.right = right; *     } * } */class Solution {
       public int minDiffInBST(TreeNode root) {
   //先中序遍历成升序数组，再从头计算相邻最小值        int min = Integer.MAX_VALUE;        List
   
     list = new ArrayList<>();        LDR(root, list);        for(int i = 0; i < list.size() - 1; i++){
               if((list.get(i+1) - list.get(i)) < min) min = list.get(i+1) - list.get(i);        }        return min;    }    public void LDR(TreeNode root,List
    
      list){
           if(root.left != null) LDR(root.left, list);		list.add(root.val);        if(root.right != null) LDR(root.right, list);    }}
    
   

解法二：

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode() {} *     TreeNode(int val) { this.val = val; } *     TreeNode(int val, TreeNode left, TreeNode right) { *         this.val = val; *         this.left = left; *         this.right = right; *     } * } */class Solution {
       public int min = Integer.MAX_VALUE;    public TreeNode pre = null;    public int minDiffInBST(TreeNode root) {
   //在中序遍历过程中记录前一个节点，直接计算这两个节点的值，取最小值        pre = new TreeNode(Integer.MAX_VALUE);        LDR(root);        return min;    }    public void LDR(TreeNode root){
           if(root.left != null) LDR(root.left);        min = Math.min(Math.abs(root.val-pre.val),min);        pre = root;        if(root.right != null) LDR(root.right);    }}

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