-
Notifications
You must be signed in to change notification settings - Fork 0
/
MinimumAbsoluteDifferenceInBST.java
54 lines (48 loc) · 1.35 KB
/
MinimumAbsoluteDifferenceInBST.java
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
package easy;
/**
* 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.
*
* 解决思路:
* 嵌套DFS搜索比较差的绝对值
* 注意只和当前节点以下节点比较,当前节点与它之上的节点肯定已经被比较过了
*
* Created by second on 2017/9/22.
*/
public class MinimumAbsoluteDifferenceInBST {
int min = Integer.MAX_VALUE;
TreeNode currentNode;
public int getMinimumDifference(TreeNode root) {
if (root == null) return min;
currentNode = root;
search(root);
getMinimumDifference(root.left);
getMinimumDifference(root.right);
return min;
}
private void search(TreeNode rootNode){
if (rootNode == null) return;
int diff = Math.abs(currentNode.val - rootNode.val);
if (rootNode != currentNode && diff < min) min = diff;
search(rootNode.left);
search(rootNode.right);
}
private class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode(int x) { val = x; }
}
}