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# Minimum Distance Between BST Nodes LeetCode Solution

## Problem – Minimum Distance Between BST Nodes

Given the `root` of a Binary Search Tree (BST), return the minimum difference between the values of any two different nodes in the tree.

Example 1:

``````Input: root = [4,2,6,1,3]
Output: 1``````

Example 2:

``````Input: root = [1,0,48,null,null,12,49]
Output: 1``````

Constraints:

• The number of nodes in the tree is in the range `[2, 100]`.
• `0 <= Node.val <= 105`

### Minimum Distance Between BST Nodes LeetCode Solution in C++

``````class Solution {
public:
int res = INT_MAX, pre = -1;
int minDiffInBST(TreeNode* root) {
if (root->left != NULL) minDiffInBST(root->left);
if (pre >= 0) res = min(res, root->val - pre);
pre = root->val;
if (root->right != NULL) minDiffInBST(root->right);
return res;
}
};``````

### Minimum Distance Between BST Nodes LeetCode Solution in Java

``````class Solution {
Integer res = Integer.MAX_VALUE, pre = null;
public int minDiffInBST(TreeNode root) {
if (root.left != null) minDiffInBST(root.left);
if (pre != null) res = Math.min(res, root.val - pre);
pre = root.val;
if (root.right != null) minDiffInBST(root.right);
return res;
}
}
``````

### Minimum Distance Between BST Nodes LeetCode Solution in Python

``````class Solution(object):
pre = -float('inf')
res = float('inf')

def minDiffInBST(self, root):
if root.left:
self.minDiffInBST(root.left)
self.res = min(self.res, root.val - self.pre)
self.pre = root.val
if root.right:
self.minDiffInBST(root.right)
return self.res``````
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