530. Minimum Absolute Difference in BST

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.

求二叉搜索树中，任意两个节点的值之间的最小绝对差值

C++（19ms）：

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     int getMinimumDifference(TreeNode* root) {
13         int minValue = INT_MAX ;
14         int val = -1 ;
15         inorder(root,minValue,val) ;
16         return minValue ;
17     }
18     
19     void inorder(TreeNode* root , int& minValue , int& val){
20         if (root->left != NULL)
21             inorder(root->left,minValue,val) ;
22         if (val >= 0)
23             minValue = min(minValue , root->val - val) ;
24         val = root->val ;
25         if (root->right != NULL)
26             inorder(root->right,minValue,val) ;
27     }
28 };