1. 程式人生 > >【LeetCode】783. Minimum Distance Between BST Nodes 解題報告(Python)

【LeetCode】783. Minimum Distance Between BST Nodes 解題報告(Python)

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:

  1. The size of the BST will be between 2 and 100.
  2. 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