Leetcode 530. Minimum Absolute Difference in BST

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).

Note: There are at least two nodes in this BST.

标签 Binary Search Tree

类似题目 (E) K-diff Pairs in an Array

思路：迭代后序遍历平衡二叉树，得到左子树的最小绝对值差lmin和左子树最大值maxl，得到右子树的最小绝对值差rmin和右子树最小值minr，然后返回min（lmin,rmin,root.val-maxl,root.val-minr）。

代码：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    TreeSet<Integer> ts = new TreeSet();
    public int getMinimumDifference(TreeNode root) {
        if (root == null)
            return Integer.MAX_VALUE;
        int lmin = getMinimumDifference(root.left);
        if (ts.floor(root.val) != null) {
            lmin = Math.min(root.val - ts.floor(root.val), lmin);
        }
        int rmin = getMinimumDifference(root.right);
        if (ts.ceiling(root.val) != null) {
            rmin = Math.min(ts.ceiling(root.val) - root.val, rmin);
        }
        ts.add(root.val);
        return Math.min(lmin, rmin);
    }
}