530. Minimum Absolute Difference in BST
阿新 • • 發佈:2018-06-07
http floor != quest inpu min 差值 question 題目
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.
解題:求二叉排序樹上任意兩個結點之間的最小差值。BST,又稱二叉排序樹,滿足“左結點的值永遠小於根結點的值,右結點的值永遠大於根結點的值”這個條件。中序遍歷所得到的序列順序,便是將結點的值從小到大排列所得到順序。那麽,這個題目就迎刃而解了,要想val值相差最小,那麽必定是中序遍歷時相鄰的兩個結點。所以在中序遍歷的過程中,保存父節點的值,計算父節點與當前結點的差值,再與min值相比較,如果比min小,則更新min,反之繼續遍歷。代碼如下:
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 class Solution {
11 private int min = Integer.MAX_VALUE;
12 private int pre = -1;//保存父節點的值
13
14 public int getMinimumDifference(TreeNode root) {
15 if(root == null )
16 return min;
17 getMinimumDifference(root.left);
18 if(pre != -1)
19 min = Math.min(min, Math.abs(root.val - pre));
20 pre = root.val;
21 getMinimumDifference(root.right);
22 return min;
23 }
24 }
也有其他的方法,比如通過java中的排序樹來做,代碼如下:
public class Solution {
TreeSet<Integer> set = new TreeSet<>();
int min = Integer.MAX_VALUE;
public int getMinimumDifference(TreeNode root) {
if (root == null) return min;
if (!set.isEmpty()) {
if (set.floor(root.val) != null) {
min = Math.min(min, root.val - set.floor(root.val));
}
if (set.ceiling(root.val) != null) {
min = Math.min(min, set.ceiling(root.val) - root.val);
}
}
set.add(root.val);
getMinimumDifference(root.left);
getMinimumDifference(root.right);
return min;
}
}
TreeSet中的 floor( ) 函數能返回小於等於給定元素的最大值, ceiling() 函數能返回大於等於給定元素的最小值,其時間開銷為對數級,還是挺快的。
參考博客:https://www.cnblogs.com/zyoung/p/6701364.html
530. Minimum Absolute Difference in BST