530.Minimum Absolute Difference in BST 二叉搜索树中的最小差的绝对值

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).

 

复习了还是不会的地方：不知道怎么遍历left right的每一个节点。
其实就指定一下root = prev就行了，dfs遍历的过程会自动遍历到每个节点

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    int min = Integer.MAX_VALUE;
    TreeNode prev = null;
    
    public int getMinimumDifference(TreeNode root) {
        //corner case
        if (root == null) {
            return min;
        }
        //in-order traversal
        getMinimumDifference(root.left);
        if (prev != null) {//only deletable if not null
            min = Math.min(min, root.val - prev.val);
        }
        //refresh the prev
        prev = root;
        getMinimumDifference(root.right);
        //return
        return min;
    }
}