LeetCode | 0530. 二叉搜索树的最小绝对差【Python】
Problem
Given a binary search tree with non-negative values, find the minimum absolute difference between values of any two nodes.
Example:
Input:
1
\
3
/
2
Output:
1
Explanation:
The minimum absolute difference is 1, which is the difference between 2 and 1 (or between 2 and 3).Note:
- There are at least two nodes in this BST.
- This question is the same as 783: https://leetcode.com/problems/minimum-distance-between-bst-nodes/
问题
给你一棵所有节点为非负值的二叉搜索树，请你计算树中任意两节点的差的绝对值的最小值。
示例：
输入：
1
\
3
/
2
输出：
1
解释：
最小绝对差为 1，其中 2 和 1 的差的绝对值为 1（或者 2 和 3）。提示：
- 树中至少有 2 个节点。
- 本题与 783 https://leetcode-cn.com/problems/minimum-distance-between-bst-nodes/ 相同
思路
DFS
法一：dfs遍历取节点值，再单独计算最小绝对差 法二：dfs遍历直接进行绝对值比较
Python3 代码
法一
# 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:
# solution one: dfs遍历取节点值，再单独计算最小绝对差
def dfs(root):
if not root:
return
# 中序遍历是递增的
if root.left:
dfs(root.left)
tmp_val.append(root.val)
if root.right:
dfs(root.right)
tmp_val = []
dfs(root)
res = float("inf")
for i in range(len(tmp_val) - 1):
res = min(res, abs(tmp_val[i] - tmp_val[i + 1]))
return res 法二
# 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:
# solution two: dfs遍历直接进行绝对值比较
pre = -1
res = float("inf")
def dfs(root):
nonlocal pre, res
if not root:
return
# 中序遍历是递增的
if root.left:
dfs(root.left)
if pre != -1:
res = min(res, abs(pre - root.val))
pre = root.val
if root.right:
dfs(root.right)
dfs(root)
return res GitHub 链接
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