LeetCode-530. Minimum Absolute Difference in BST (Java)
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Given a binary search tree with non-negative values, find the minimum absolute difference between values of any two nodes.
Example:
Input: 1 \ 3 / 2Output:1Explanation:The minimum absolute difference is 1, which is the difference between 2 and 1 (or between 2 and 3).
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这道题是关于二叉搜索树,在解题时,我未注意a binary search tree,虽然测试通过,但耗时过多,在此基础上,求最小值的时候,我每个数与所有的数都做了差,如果注意到二叉搜索树并且了解其性质,则只需要求其相邻的数之差即可。
关于二叉搜索树
若它的左子树不为空,则左子树上所有结点的值均小于等于它的根结点的值;
若它的右子树不为空,则右子树上所有结点的值均大于等于它的根结点的值;
它的左、右子树也分别为二叉查找树。
目的并非为了排序,而是为了提高查找和删除关键字的速度,不管怎么说,在一个有序的数据集上查找,
速度总是要快于无序的数据集。
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代码如下:
public class Solution { int min = Integer.MAX_VALUE; Integer prev = null; public int getMinimumDifference(TreeNode root) { if (root == null) return min; getMinimumDifference(root.left); if (prev != null) { min = Math.min(min, root.val - prev); } prev = root.val; getMinimumDifference(root.right); return min; }}
二叉搜索树使用中序遍历,得到的结果是已经排序好的,所以只需比较相邻的值即可。
思路如下:求最小差----->对数字进行排序后,相邻做差------->二叉搜索树,中序遍历可得已排序序列。
参考链接:https://discuss.leetcode.com/topic/80823/two-solutions-in-order-traversal-and-a-more-general-way-using-treeset
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