【LeetCode】783. Minimum Distance Between BST Nodes 解題報告(Python)
阿新 • • 發佈:2019-02-16
目錄
題目描述
Given a Binary Search Tree (BST) with the root node root, return the minimum difference between the values of any two different nodes in the tree.
Example : Input: root = [4,2,6,1,3,null,null] Output: 1 Explanation: Note that root is a TreeNode object, not an array. The given tree [4,2,6,1,3,null,null] is represented by the following diagram: 4 / \ 2 6 / \ 1 3 while the minimum difference in this tree is 1, it occurs between node 1 and node 2, also between node 3 and node 2.
Note:
- The size of the BST will be between 2 and 100.
- The BST is always valid, each node’s value is an integer, and each node’s value is different.
題目大意
求BST的兩個節點之間的最小差值。
解題方法
中序遍歷
看見BST想中序遍歷是有序的啊~所以先進性中序遍歷,得到有序列表,然後找出相鄰的兩個節點差值的最小值即可。
# 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):
"""
:type root: TreeNode
:rtype: int
"""
vals = []
def inOrder(root):
if not root:
return
inOrder(root.left)
vals.append(root.val)
inOrder(root.right)
inOrder(root)
return min([vals[i + 1] - vals[i] for i in xrange(len(vals) - 1)])
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def minDiffInBST(self, root):
"""
:type root: TreeNode
:rtype: int
"""
self.res = float("inf")
self.prev = None
self.inOrder(root)
return self.res
def inOrder(self, root):
if not root: return
self.inOrder(root.left)
if self.prev:
self.res = min(self.res, root.val - self.prev.val)
self.prev = root
self.inOrder(root.right)
日期
2018 年 2 月 28 日
2018 年 11 月 14 日 —— 很嚴重的霧霾