LeetCode Minimum Absolute Difference in BST

原题链接在这里：https://leetcode.com/problems/minimum-absolute-difference-in-bst/#/description

题目：

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.

题解：

Binary Tree Inorder Traversal, 由于是BST, 所以是asending的, 取出最小difference.

Time Complexity: O(n). Space: O(logn), stack space.

AC Java:

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     int min = Integer.MAX_VALUE;
12     Integer pre = null;
13     public int getMinimumDifference(TreeNode root) {
14         if(root == null){
15             return min;
16         }
17         getMinimumDifference(root.left);
18         
19         if(pre != null){
20             min = Math.min(min, root.val-pre);
21         }
22         pre = root.val;
23         
24         getMinimumDifference(root.right);
25         return min;
26     }
27 }

如果不是BST的话可以借助于TreeSet<Integer> ts, 对于每一个node, 找出node.val在ts中的floor和ceil, 计算minimum difference. 再把node 本身的val加到ts中.

Time Complexity: O(nlogn). Space: O(n), ts size.

AC Java:

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     int min = Integer.MAX_VALUE;
12     TreeSet<Integer> ts = new TreeSet<Integer>();
13     
14     public int getMinimumDifference(TreeNode root) {
15         if(root == null){
16             return min;
17         }
18         
19         if(!ts.isEmpty()){
20             if(ts.floor(root.val) != null){
21                 min = Math.min(min, root.val-ts.floor(root.val));
22             }
23             if(ts.ceiling(root.val) != null){
24                 min = Math.min(min, ts.ceiling(root.val)-root.val);
25             }
26         }
27         ts.add(root.val);
28         
29         getMinimumDifference(root.left);
30         getMinimumDifference(root.right);
31         return min;
32     }
33 }