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530. Minimum Absolute Difference in BST

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Description

Given the root of a Binary Search Tree (BST), return the minimum absolute 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, 104].
• 0 <= Node.val <= 105

 

Note: This question is the same as 783: https://leetcode.com/problems/minimum-distance-between-bst-nodes/

Solutions

Python3

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def getMinimumDifference(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

Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {

    private int minDiff = Integer.MAX_VALUE;
    private Integer pre;

    public int getMinimumDifference(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);
    }
}

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