方法一,非递归方法,中序遍历

 /**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/ static int wing=[]()
{
std::ios::sync_with_stdio(false);
cin.tie(NULL);
return ;
}(); class Solution
{
public: int minDiffInBST(TreeNode* root)
{
stack<TreeNode*>s;
int mindiff=INT_MAX;
TreeNode *p=root;
int pre=-(root->val),cur=;
while(p||!s.empty())
{
if(p)
{
s.push(p);
p=p->left;
}
else
{
p=s.top();
s.pop();
cur=p->val;
mindiff=min(mindiff,cur-pre);
pre=cur;
p=p->right;
}
}
return mindiff;
}
};

方法二,递归方法,中序遍历

 /**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/ static int wing=[]()
{
std::ios::sync_with_stdio(false);
cin.tie(NULL);
return ;
}(); class Solution {
int dif=INT_MAX, val=-;
public:
int minDiffInBST(TreeNode* root)
{
if(root->left!=nullptr)
minDiffInBST(root->left);
if(val>=) dif=min(dif,root->val-val);
val=root->val;
if(root->right!=nullptr)
minDiffInBST(root->right);
return dif;
} };

简单,问题不大

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