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Leetcode 530. Minimum Absolute Difference in BST

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Question

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

Example: 

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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

Explanation

按照之前Leetcode 94.Binary Tree Inorder Traversal的思路，可用递归来做。但是如果不是BST的话，我们也可以用Treeset来存值，同样用递归。 两种方法如下，第一种Time complexity O(N), space complexity O(1)，第二种的话Time complexity O(NlgN), space complexity O(N).

Code

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    // Recursive:
    Time complexity O(N), space complexity O(1).
    int min = Integer.MAX_VALUE;
    TreeNode pre = null;
    public int getMinimumDifference(TreeNode root) {
        if (root == null) return 0;
        getMinimumDifference(root.left);
        if (pre != null) {
            min = Math.min(min, root.val - pre.val);
        }
        pre = root;
        getMinimumDifference(root.right);
        return min;
    }
    // Another Recursive
    // If the tree is not BST, we can just use TreeSet to store all the nodes
    TreeSet<Integer> set = new TreeSet<>();
    int min = Integer.MAX_VALUE;
    public int getMinimumDifference(TreeNode root) {
        if (root == null) return 0;
        if (!set.isEmpty()) {
            if (set.ceiling(root.val)!=null) {
                min = Math.min(min, Math.abs(root.val - set.ceiling(root.val)));
            }
            if (set.floor(root.val)!=null) {
                min = Math.min(min, Math.abs(root.val - set.floor(root.val)));
            }
        }
        set.add(root.val);
        getMinimumDifference(root.left);
        getMinimumDifference(root.right);
        return min;
    }
}