【发布时间】:2020-01-21 02:38:12
【问题描述】:
我正在解决 Leetcode 问题,可以在这里找到:https://leetcode.com/problems/minimum-distance-between-bst-nodes/
问题: 给定一个具有根节点的二叉搜索树 (BST),返回树中任意两个不同节点的值之间的最小差值。
例子:
输入:root = [4,2,6,1,3,null,null] 输出:1 解释: 请注意,root 是 TreeNode 对象,而不是数组。
给定的树[4,2,6,1,3,null,null]用下图表示:
4
/ \
2 6
/ \
1 3
虽然这棵树的最小差值为 1,但它发生在节点 1 和节点 2 之间,也发生在节点 3 和节点 2 之间。
注意:我已经实现了前序遍历,但是我的代码遇到了堆栈溢出错误,谁能帮忙指出逻辑错误在哪里?
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
def minDiffInBST(self, root):
min_value = float('inf')
def helper(node, min_value):
print(node.val, "and", min_value)
# if root is None
if not root:
return None
if node.left:
min_value = min(min_value, node.val - node.left.val)
if node.right:
min_value = min(min_value, node.right.val - node.val)
helper(root.left, min_value)
helper(root.right, min_value)
return min_value
helper(root, min_value)
更改后:
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
def minDiffInBST(self, root):
min_value = float('inf')
def helper(node, min_value):
# if root is None
if not node:
return node
print(node.val, min_value)
if (node.left):
min_value = min(min_value, node.val - node.left.val)
if (node.right):
min_value = min(min_value, node.right.val - node.val)
helper(node.left, min_value)
helper(node.right, min_value)
return min_value
helper(root, min_value)
【问题讨论】:
标签: python recursion binary-search-tree depth-first-search