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.
解题:求二叉排序树上任意两个结点之间的最小差值。BST,又称二叉排序树,满足“左结点的值永远小于根结点的值,右结点的值永远大于根结点的值”这个条件。中序遍历所得到的序列顺序,便是将结点的值从小到大排列所得到顺序。那么,这个题目就迎刃而解了,要想val值相差最小,那么必定是中序遍历时相邻的两个结点。所以在中序遍历的过程中,保存父节点的值,计算父节点与当前结点的差值,再与min值相比较,如果比min小,则更新min,反之继续遍历。代码如下:
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;//保存父节点的值 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,比如通过java中的排序树来做,代码如下:
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( ) 函数能返回小于等于给定元素的最大值, ceiling() 函数能返回大于等于给定元素的最小值,其时间开销为对数级,还是挺快的。
参考博客:https://www.cnblogs.com/zyoung/p/6701364.html
原文地址:https://www.cnblogs.com/phdeblog/p/9149547.html
时间: 2024-10-09 12:38:58