# 783. Minimum Distance Between BST Nodes

## Description

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

Note: This question is the same as 530: https://leetcode.com/problems/minimum-absolute-difference-in-bst/

## Solutions

• /**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode() {}
*     TreeNode(int val) { this.val = val; }
*     TreeNode(int val, TreeNode left, TreeNode right) {
*         this.val = val;
*         this.left = left;
*         this.right = right;
*     }
* }
*/
class Solution {
private int ans;
private int prev;
private int inf = Integer.MAX_VALUE;

public int minDiffInBST(TreeNode root) {
ans = inf;
prev = inf;
dfs(root);
return ans;
}

private void dfs(TreeNode root) {
if (root == null) {
return;
}
dfs(root.left);
ans = Math.min(ans, Math.abs(root.val - prev));
prev = root.val;
dfs(root.right);
}
}

• /**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode() : val(0), left(nullptr), right(nullptr) {}
*     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
*     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
const int inf = INT_MAX;
int ans;
int prev;

int minDiffInBST(TreeNode* root) {
ans = inf, prev = inf;
dfs(root);
return ans;
}

void dfs(TreeNode* root) {
if (!root) return;
dfs(root->left);
ans = min(ans, abs(prev - root->val));
prev = root->val;
dfs(root->right);
}
};

• # Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
def minDiffInBST(self, root: Optional[TreeNode]) -> int:
def dfs(root):
if root is None:
return
dfs(root.left)
nonlocal ans, prev
ans = min(ans, abs(prev - root.val))
prev = root.val
dfs(root.right)

ans = prev = inf
dfs(root)
return ans


• /**
* Definition for a binary tree node.
* type TreeNode struct {
*     Val int
*     Left *TreeNode
*     Right *TreeNode
* }
*/
func minDiffInBST(root *TreeNode) int {
inf := 0x3f3f3f3f
ans, prev := inf, inf
var dfs func(*TreeNode)
dfs = func(root *TreeNode) {
if root == nil {
return
}
dfs(root.Left)
ans = min(ans, abs(prev-root.Val))
prev = root.Val
dfs(root.Right)
}
dfs(root)
return ans
}

func abs(x int) int {
if x < 0 {
return -x
}
return x
}

• var minDiffInBST = function (root) {
let ans = Number.MAX_SAFE_INTEGER,
prev = Number.MAX_SAFE_INTEGER;
const dfs = root => {
if (!root) {
return;
}
dfs(root.left);
ans = Math.min(ans, Math.abs(root.val - prev));
prev = root.val;
dfs(root.right);
};
dfs(root);
return ans;
};