leetcode 530. Minimum Absolute Difference in BST（二叉搜索数中最小差的绝对值）

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).

找出二叉搜索树中两两节点间差的绝对值的最小值

思路：
二叉搜索树有个特点就是它的中序遍历值是升序的
相当于一个升序数组，两个值的最小差只能在相邻元素中产生
因为升序数组[a, b, c], 那么a < b < c, 那么c - a > b - a

可以中序遍历BST，把遍历过的节点值保存在数组中，然后数组依次两两元素求差取最小

   public int getMinimumDifference(TreeNode root) {
        List<Integer> nodes = new ArrayList<>();
        int diff = 0;
        midTraversal(root, nodes);
        
        diff = nodes.get(1) - nodes.get(0);
        
        for(int i = 2; i < nodes.size(); i++) {
            diff = Math.min(diff, nodes.get(i) - nodes.get(i-1));
        }
        return diff;
    }
    
    public void midTraversal(TreeNode root, List<Integer> nodes) {
        if(root == null) return;
        midTraversal(root.left, nodes);
        nodes.add(root.val);
        midTraversal(root.right, nodes);
    }

因为求差时只需要用到前一个值，因此可以每次访问到root时保存前一个的值，下次再求差的绝对值取最小
因为BST中值都是非负，所以pre的初始值可以取-1

class Solution {   
    int pre = -1;
    int diff = Integer.MAX_VALUE;
    public int getMinimumDifference(TreeNode root) {        
        midTraversal(root);      
        return diff;
    }
    
    public void midTraversal(TreeNode root) {
        if(root == null) return;
        midTraversal(root.left);
        if(pre != -1) {
            diff = Math.min(diff, Math.abs(root.val - pre));
        }
        pre = root.val;
        midTraversal(root.right);
    }
}