leetcode 530. Minimum Absolute Difference in BST 深度优先遍历DFS

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 <set>#include <queue>#include <stack>#include <string>#include <climits>#include <algorithm>#include <sstream>#include <functional>#include <bitset>#include <numeric>#include <cmath>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);        sort(res.begin(), res.end());        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        {            res.push_back(root->val);            dfs(root->left, res);            dfs(root->right,res);        }    }};