Given the root of a Binary Search Tree (BST), return the minimum difference between the values of any two different nodes in the tree.
Note: This question is the same as 530: https://leetcode.com/problems/minimum-absolute-difference-in-bst/
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
# 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: TreeNode) -> int:
def inorder(root):
if not root:
return
inorder(root.left)
if self.pre is not None:
self.min_diff = min(self.min_diff, abs(root.val - self.pre))
self.pre = root.val
inorder(root.right)
self.pre = None
self.min_diff = 10 ** 5
inorder(root)
return self.min_diff/**
* 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 minDiff = Integer.MAX_VALUE;
private Integer pre;
public int minDiffInBST(TreeNode root) {
inorder(root);
return minDiff;
}
private void inorder(TreeNode root) {
if (root == null) return;
inorder(root.left);
if (pre != null) minDiff = Math.min(minDiff, Math.abs(root.val - pre));
pre = root.val;
inorder(root.right);
}
}