标签：math his 替换 java nbsp mdi 遍历二叉树 note add
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.
题意及分析：给出一颗二叉搜索树（节点为非负），要求求出任意两个点之间差值绝对值的最小值。题目简单，直接中序遍历二叉树，得到一个升序排列的list，然后计算每两个数差的绝对值，每次和当前最小值进行比较，若小于当前最小值，替换掉即可。
代码：
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public int getMinimumDifference(TreeNode root) { //中根序遍历，然后比较每相邻的两个数
List<Integer> list = new ArrayList<>();
search(list,root);
int min = Integer.MAX_VALUE;
for(int i=0;i<list.size()-1;i++){
int temp = Math.abs(list.get(i+1)-list.get(i));
if(temp<min)
min = temp;
}
return min;
}
private void search( List<Integer> list,TreeNode node){
if(node!=null){
search(list,node.left);
list.add(node.val);
search(list,node.right);
}
}
}
[LeetCode] 530. Minimum Absolute Difference in BST Java
标签：math his 替换 java nbsp mdi 遍历二叉树 note add
原文地址：http://www.cnblogs.com/271934Liao/p/7522840.html
