# 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.

``````class Solution {
public:
int getMinimumDifference(TreeNode* root) {
int res = INT_MAX, pre = -1;
inorder(root, pre, res);
return res;
}
void inorder(TreeNode* root, int& pre, int& res) {
if (!root) return;
inorder(root->left, pre, res);
if (pre != -1) res = min(res, root->val - pre);
pre = root->val;
inorder(root->right, pre, res);
}
};
``````

``````class Solution {
public:
int getMinimumDifference(TreeNode* root) {
int res = INT_MAX;
helper(root, INT_MIN, INT_MAX, res);
return res;
}
void helper(TreeNode* root, int low, int high, int& res) {
if (!root) return;
if (low != INT_MIN) res = min(res, root->val - low);
if (high != INT_MAX) res = min(res, high - root->val);
helper(root->left, low, root->val, res);
helper(root->right, root->val, high, res);
}
};
``````

``````class Solution {
public:
int getMinimumDifference(TreeNode* root) {
int res = INT_MAX, pre = -1;
stack<TreeNode*> st;
TreeNode *p = root;
while (p || !st.empty()) {
while (p) {
st.push(p);
p = p->left;
}
p = st.top(); st.pop();
if (pre != -1) res = min(res, p->val - pre);
pre = p->val;
p = p->right;
}
return res;
}
};
``````

https://discuss.leetcode.com/topic/80896/my-solution-using-no-recursive-in-order-binary-tree-iteration

https://discuss.leetcode.com/topic/80823/two-solutions-in-order-traversal-and-a-more-general-way-using-treeset/2

https://discuss.leetcode.com/topic/80916/java-no-in-order-traverse-solution-just-pass-upper-bound-and-lower-bound

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