ZOJ Problem Set - 1097 Code the Tree Time Limit: 2 Seconds      Memory Limit: 65536 KB A tree (i.e. a connected graph without cycles) with vertices numbered by the integers 1, 2, ..., n is given. The "Prufer" code of such a tree is built as foll…

Code the Tree Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 2350   Accepted: 906 Description A tree (i.e. a connected graph without cycles) with vertices numbered by the integers 1, 2, ..., n is given. The "Prufer" code of such a…

Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 2292   Accepted: 878 Description A tree (i.e. a connected graph without cycles) with vertices numbered by the integers 1, 2, ..., n is given. The "Prufer" code of such a tree is built…

Code the Tree Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 2259   Accepted: 859 Description A tree (i.e. a connected graph without cycles) with vertices numbered by the integers 1, 2, ..., n is given. The "Prufer" code of such a…

G.Code the Tree Time Limit: 2 Sec  Memory Limit: 128 MB Submit: 35  Solved: 18 [Submit][Status][Web Board] Description A tree (i.e. a connected graph without cycles) with vertices numbered by the integers 1, 2, ..., n is given. The "Prufer" code…

语言:Python 描述:使用递归实现 # Definition for a binary tree node # class TreeNode: # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Solution: # @param p, a tree node # @param q, a tree node # @return a boolean def isSameTre…

题目大意:2567是给出一棵树,让你求出它的Prufer序列.2568时给出一个Prufer序列,求出这个树. 思路:首先要知道Prufer序列.对于随意一个无根树,每次去掉一个编号最小的叶子节点,并保存这个节点所连接的节点所得到的序列就是这棵树的Prufer序列. 这个序列有十分优雅的性质.它能与无根树一一相应.因此.两个标号一样的无根树得到的Prufer序列也一定是一样的. 此外,设一个节点的度数是d[i],那么他会在Prufer序列中出现d[i] - 1次. 2567:记录每个节点的度.依…

描述:递归调用,getMax返回 [节点值,经过节点左子节点的最大值,经过节点右节点的最大值],每次递归同时查看是否存在不经过节点的值大于max. 代码:待优化 def getLargeNode(self, a, b): if a and b: return max(a, b) elif a and not b: return a elif not a and b: return b else: tmp = None def getMax(self, node): if node is None…

语言:Python 描述:使用递归实现 class Solution: # @return an integer def numTrees(self, n): : elif n == : else: part_1 = self.numTrees(n-) * part_2 = ,n-): part_left = self.numTrees(i) part_right = self.numTrees(n - - i) part_2 += part_left * part_right return p…

Experiment report of Besti course:<Program Design & Data Structures> Class: 1623 Student Name: Wang, Yixiao Student Number:20162314 Tutor:Mr.Lou.Mr.Wang Experiment date:2017.10.27 Secret level: Unsecretive Experiment time:60 minutes Major/Electi…