# 530.Minimum Absolute Difference in BST

**530.Minimum Absolute Difference in BST**

难度:Easy

> 给定一个所有节点为非负值的二叉搜索树，求树中任意两节点的差的绝对值的最小值。

示例 :

```
输入:

   1
    \
     3
    /
   2

输出:
1
```

解释: 最小绝对差为1，其中 2 和 1 的差的绝对值为 1（或者 2 和 3）。 注意: 树中至少有2个节点。

递归判断，当前根节点的最小绝对差是与其左子树的最后最右节点的绝对差，和其与右子树的最后最左节点的绝对差的较小值。

```
/**
 * 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) {
           int m=std::numeric_limits<int>::max();
        if(!root) return m;
        int ld=m,rd=m;
        if(root->left)
        {
            TreeNode* leftNode=root->left;
            while(leftNode)
            {
                ld=root->val - leftNode->val;
                leftNode=leftNode->right;
            }
            
            
        }
        if(root->right)
        {
            TreeNode* rightNode=root->right;
            while(rightNode)
            {
                rd=rightNode->val-root->val;
                rightNode=rightNode->left;
            }
            
            
        }

     
      //  cout<<ld << " " <<rd << " "<< m<<endl;
        if(root->left&& root->right) 
        {
            m=min(ld,rd);
            m=min(m,min(getMinimumDifference(root->left),getMinimumDifference(root->right)));
        }
        else if (root->left) 
            m=min(ld,getMinimumDifference(root->left));
        else if(root->right) m=min(rd,getMinimumDifference(root->right));
        else 
            return m;
      //  cout<<m<<endl;
        return m;
    }
};
```

> 执行用时 : 40 ms, 在Minimum Absolute Difference in BST的C++提交中击败了86.97% 的用户\
> 内存消耗 : 21.8 MB, 在Minimum Absolute Difference in BST的C++提交中击败了90.12% 的用户


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