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.

题意很简单就是一个简单的DFS深度优先遍历，

代码如下：

#include <iostream>
#include <vector>
#include <map>
#include <unordered_map>
#include <set>
#include <unordered_set>
#include <queue>
#include <stack>
#include <string>
#include <climits>
#include <algorithm>
#include <sstream>
#include <functional>
#include <bitset>
#include <numeric>
#include <cmath>
#include <regex>
#include <iomanip>

using namespace std;


/*
struct TreeNode
{
int val;
TreeNode *left;
TreeNode *right;
TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
*/


class Solution
{
public:
    int getMinimumDifference(TreeNode* root)
    {
        vector<int> res;
        dfs(root, res);

        int minDif = INT_MAX;
        for (int i = 1; i < res.size(); i++)
            minDif = min(minDif, res[i] - res[i - 1]);
        return minDif;
    }

    void dfs(TreeNode* root, vector<int>& res)
    {
        if (root == NULL)
            return;
        else
        {
            dfs(root->left, res);
            res.push_back(root->val);
            dfs(root->right, res);
        }
    }
};