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/

Submitted Code

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right

class Solution(object):
    def minDiffInBST(self, root):
        """
        :type root: Optional[TreeNode]
        :rtype: int
        """
        self.prev = None
        self.min_diff = float('inf')
        
        def inorder(node):
            if node:
                inorder(node.left)
                if self.prev is not None:
                    self.min_diff = min(self.min_diff, (node.val - self.prev))
                self.prev = node.val
                inorder(node.right)

        inorder(root)
        return self.min_diff

Runtime: 0 ms | Beats 100.00%
Memory: 12.45 MB | Beats 96.41%

530번과 동일한 문제로, 중위 순회로 정렬된 결과에서 앞뒤 노드끼리의 차이를 비교한다.

Other Solutions

1st

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

time complexity: 𝑂(𝑛)
space complexity: 𝑂(ℎ)

이전에 방문한 노드의 값을 저장하는 pre의 값을 None이 아니라 음의 무한대로 설정하면 더 간단하게 코드를 짤 수 있다.