题目来源
求二叉搜索树的任意两个节点间的最小的差值，注意利用其性质，中序遍历就是一个递增序列，然后计算前后两个节点的差值就可以了，代码如下：
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
int getMinimumDifference(TreeNode* root) {
int last = -1, minimum = INT_MAX;
if (root == NULL)
return 0;
dfs(root, last, minimum);
return minimum;
}
void dfs(TreeNode *node, int &last, int &minimum)
{
if (node->left)
dfs(node->left, last, minimum);
if (last >= 0)
minimum = min(minimum, abs(node->val-last));
last = node->val;
if (node->right)
dfs(node->right, last, minimum);
}
};
