530. Minimum Absolute Difference in BST

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    /   2Output:1Explanation:The minimum absolute difference is 1, which is the difference between 2 and 1 (or between 2 and 3).

思路：

本题思路比较简单，求任意结点间的差值绝对值，可以先利用中序遍历存储结点，因为中序遍历的结点是按升序排列的，所以只要比较相邻点的差值即可

代码：

/** * 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) {        vector<int> res;        InOrder(root,res);        int min =res.back(),diff;        for(int i = 1;i<res.size();i++)        {            diff = res[i] - res[i-1];            min = min < diff?min:diff;        }        return min;            }private:    void InOrder(TreeNode* T,vector<int>& res){      if(T != NULL){          //访问左子结点          InOrder(T->left,res);          //访问根节点          res.push_back(T->val);        //访问右子结点          InOrder(T->right,res);      }  }};