Formatted question description: https://leetcode.ca/all/783.html

# 783. Minimum Distance Between BST Nodes (Medium)

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

Example :

Input: root = [4,2,6,1,3,null,null]
Output: 1
Explanation:
Note that root is a TreeNode object, not an array.

The given tree [4,2,6,1,3,null,null] is represented by the following diagram:

      4
/   \
2      6
/ \
1   3


while the minimum difference in this tree is 1, it occurs between node 1 and node 2, also between node 3 and node 2.

Note:

1. The size of the BST will be between 2 and 100.
2. The BST is always valid, each node’s value is an integer, and each node’s value is different.

## Solution 1.

// OJ: https://leetcode.com/problems/minimum-distance-between-bst-nodes/

// Time: O(N)
// Space: O(log(N))
class Solution {
private:
int ans = INT_MAX;
TreeNode *prev = NULL;
void inorder(TreeNode *root) {
if (!root) return;
inorder(root->left);
if (prev) ans = min(ans, root->val - prev->val);
prev = root;
inorder(root->right);
}
public:
int minDiffInBST(TreeNode* root) {
inorder(root);
return ans;
}
};


Java

/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
class Solution {
public int minDiffInBST(TreeNode root) {
List<Integer> inorderTraversal = inorderTraversal(root);
int minDiff = Integer.MAX_VALUE;
int size = inorderTraversal.size();
for (int i = 1; i < size; i++) {
int diff = inorderTraversal.get(i) - inorderTraversal.get(i - 1);
minDiff = Math.min(minDiff, diff);
}
return minDiff;
}

public List<Integer> inorderTraversal(TreeNode root) {
List<Integer> inorderTraversal = new ArrayList<Integer>();
Stack<TreeNode> stack = new Stack<TreeNode>();
TreeNode node = root;
while (!stack.isEmpty() || node != null) {
while (node != null) {
stack.push(node);
node = node.left;
}
TreeNode visitNode = stack.pop();