Leetcode530. Minimum Absolute Difference in BST

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本文针对LeetCode上的第530题——求二叉搜索树中两节点间的最小绝对差，提供了详细的解题思路及代码实现。通过递归遍历二叉树的方法，寻找左右子树的最大最小值来确定最小绝对差。

Leetcode530. 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
    /
   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.

题目分析：

首先从题意分析，这是一颗二叉搜索树，当你处于根节点的时候，要和左边最大的比较，和右边最小的比较，同时设一个全局变量min，然后不断左递归和右递归。

代码：

class Solution {
private:
     int min1 = INT_MAX - 1;
public:
    int getMinimumDifference(TreeNode* root) {
        getMini(root);
        return min1;
    }

    void getMini(TreeNode* root) {
        if (!root) return;
        int left, right;
        if (root->left) {
            left = max(root->left);
            if (min1 > abs(root->val - left)) min1 = abs(root->val - left);
            getMini(root->left);
        }
        if (root->right) {
             right = min(root->right);
             if (min1 > abs(root->val - right)) min1 = abs(right - root->val);
            getMini(root->right);
         }
        
    }

    int min(TreeNode* root) {
        if (root == NULL) return -1;
        if (root->left == NULL) return root->val;
        return min(root->left);
    }
    int max(TreeNode* root) {
        if (root == NULL) return -1;
        if (root->right == NULL) return root->val;
        return max(root->right);
    }
};