530. Minimum Absolute Difference in BST 530. BST的最小绝对差

Given a binary search tree with non-negative values, find the minimum absolute difference between values of any two nodes.

Example:

Input:

   1
    
     3
    /
   2

Output:
1

Explanation:
The minimum absolute difference is 1, which is the difference between 2 and 1 (or between 2 and 3).


BST一般都是中序，不然也没必要出BST的题了啊

左

打印。这里怎么去和别的所有节点、包括之前的取得联系呢？这个地方我觉得挺困难的。
就是prev = root;就行了，具体的迭代不用管
右

TreeNode prev = root; 比较完了再更改

Math.abs(root.val - prev.val)); 而且要用绝对值啊！

/**
 * 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 {
    int result = Integer.MAX_VALUE;
    TreeNode prev = new TreeNode(0);
    
    public int getMinimumDifference(TreeNode root) {
        //cc
        if (root == null) {
            return 0;
        }
        
        getMinimumDifference(root.left);
        
        
        if (Math.abs(root.val - prev.val) < result) {
            result = Math.abs(root.val - prev.val);
        }
        
        prev = root;

        getMinimumDifference(root.right);
        
        return result;
    }
}