530. Minimum Absolute Difference in BST

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).Note: There are at least two nodes in this BST.

代码：

struct TreeNode {    int val;    TreeNode *left;    TreeNode *right;    TreeNode(int x) : val(x), left(NULL), right(NULL) {}};class Solution {public:    void inorderTraverse(TreeNode* root, int& val, int& min_dif) {        if (root->left != NULL)             inorderTraverse(root->left, val, min_dif);        if (val >= 0)             min_dif = min(min_dif, root->val - val);        val = root->val;        if (root->right != NULL)             inorderTraverse(root->right, val, min_dif);    }    int getMinimumDifference(TreeNode* root) {        auto min_dif = INT_MAX, val = -1;        inorderTraverse(root, val, min_dif);        return min_dif;    }};

解题思路：

• 采用递归中序遍历的思路，由于是个二叉搜索树，任意两点间差值的最小值肯定是从相邻结点间的差值中选一个；
• 令val = -1 并且设置判别条件 val >= 0 避免了叶子结点与不存在的下一层结点做差，即val - (-1)，此处的-1为val的初始值；
• 每次比较min的值与当前两个结点的差值大小，选择较小的哪一个值，赋给min；
• 返回min。