leetcode 530. Minimum Absolute Difference in BST

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本文介绍了一种通过中序遍历二叉搜索树的方法来找出树中任意两节点间最小绝对差值的有效算法。该方法利用栈结构实现了非递归的中序遍历，并在遍历过程中比较相邻节点值，从而找到最小绝对差。

530. Minimum Absolute Difference in BST

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.

利用中序遍历，前后对比

/**
 * 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) 
    {
        stack<TreeNode *> s;  
        TreeNode * p = root;  
        int ret = INT_MAX;
        int last = -1;
        while (p != NULL || !s.empty())  
        {  
            while ( p != NULL)  
            {  
                s.push(p);  
                p = p->left;  
            }  
            if (!s.empty())   //来到了最左边
            {  
                p = s.top();  
                if (last != -1)
                    ret = min(ret, p->val - last);  
                last = p->val;
                s.pop();  
                p = p->right;  
            }  
        }   
        return ret;  
    }
};