PTA数据结构与算法题目集(中文) 7-31

PTA数据结构与算法题目集(中文)  7-31

7-31 笛卡尔树 (25 分)
 

笛卡尔树是一种特殊的二叉树，其结点包含两个关键字K1和K2。首先笛卡尔树是关于K1的二叉搜索树，即结点左子树的所有K1值都比该结点的K1值小，右子树则大。其次所有结点的K2关键字满足优先队列（不妨设为最小堆）的顺序要求，即该结点的K2值比其子树中所有结点的K2值小。给定一棵二叉树，请判断该树是否笛卡尔树。

输入格式:

输入首先给出正整数N（≤1000），为树中结点的个数。随后N行，每行给出一个结点的信息，包括：结点的K1值、K2值、左孩子结点编号、右孩子结点编号。设结点从0~(N-1)顺序编号。若某结点不存在孩子结点，则该位置给出−。

输出格式:

输出YES如果该树是一棵笛卡尔树；否则输出NO。

输入样例1:

6
8 27 5 1
9 40 -1 -1
10 20 0 3
12 21 -1 4
15 22 -1 -1
5 35 -1 -1

输出样例1:

YES

输入样例2:

6
8 27 5 1
9 40 -1 -1
10 20 0 3
12 11 -1 4
15 22 -1 -1
50 35 -1 -1

输出样例2:

NO
题目分析:一道树的应用题 主要考察的是 对平衡二叉树定义 以及 优先队列(最小堆)定义的理解 对平衡二叉树判断时要注意 不仅要满足每个子树比左边小比右边大 整体树也要满足平衡二叉树的概念
  1 #define _CRT_SECURE_NO_WARNINGS
  2 #include<stdio.h>
  3 #include<string.h>
  4 #include<malloc.h>
  5 
  6 struct TreeNode
  7 {
  8     int K1;
  9     int K2;
 10     int Lc;
 11     int Rc;
 12 }Tr[1000];
 13 
 14 int Collected[1000];
 15 
 16 int FindTree(int N)
 17 {
 18     for (int i = 0; i < N; i++)
 19         if (!Collected[i])
 20             return i;
 21 }
 22 
 23 
 24 
 25 int IsAVL(int Tree)
 26 {
 27     if (Tree ==-1)
 28         return 1;
 29     else
 30     {
 31         if (Tr[Tree].Lc != -1 && Tr[Tree].Rc != -1)
 32             if (Tr[Tree].K1 >=Tr[Tr[Tree].Lc].K1 && Tr[Tree].K1 < Tr[Tr[Tree].Rc].K1)
 33                 return IsAVL(Tr[Tree].Lc) && IsAVL(Tr[Tree].Rc);
 34             else
 35                 return 0;
 36         else if (Tr[Tree].Lc == -1 && Tr[Tree].Rc == -1)
 37             return 1;
 38         else if (Tr[Tree].Lc == -1)
 39             return Tr[Tree].K1 < Tr[Tr[Tree].Rc].K1;
 40         else
 41             return Tr[Tree].K1 >=Tr[Tr[Tree].Lc].K1;
 42 
 43     }
 44 }
 45 
 46 int IsMinHeap(int Tree)
 47 {
 48     if (Tree ==-1)
 49         return 1;
 50     else
 51     {
 52         if (Tr[Tree].Lc != -1 && Tr[Tree].Rc != -1)
 53             if (Tr[Tree].K2 <=Tr[Tr[Tree].Lc].K2 && Tr[Tree].K2 <=Tr[Tr[Tree].Rc].K2)
 54                 return IsMinHeap(Tr[Tree].Lc) && IsMinHeap(Tr[Tree].Rc);
 55             else
 56                 return 0;
 57         else if (Tr[Tree].Lc == -1 && Tr[Tree].Rc == -1)
 58             return 1;
 59         else if (Tr[Tree].Lc == -1)
 60             return Tr[Tree].K2 <=Tr[Tr[Tree].Rc].K2;
 61         else
 62             return Tr[Tree].K2 <=Tr[Tr[Tree].Lc].K2;
 63     }
 64 }
 65 int JudgetLeft(int Tree, int T);
 66 int JudgetRight(int Tree, int T);
 67 
 68 int JudgetLeft(int Tree,int T)
 69 {
 70     if (T == -1 || Tree == -1)
 71         return 1;
 72     if (Tr[Tree].K1 > Tr[T].K1)
 73         return JudgetLeft(Tree, Tr[T].Lc) && JudgetLeft(Tree,Tr[T].Rc);
 74     else
 75         return 0;
 76 }
 77 
 78 int JudgetRight(int Tree, int T)
 79 {
 80     if (T == -1 || Tree == -1)
 81         return 1;
 82     if (Tr[Tree].K1 < Tr[T].K1)
 83         return JudgetRight(Tree, Tr[T].Lc) && JudgetRight(Tree, Tr[T].Rc);
 84     else
 85         return 0;
 86 }
 87 
 88 int IsTree(int Tree)
 89 {
 90     if (Tree == -1)
 91         return 1;
 92     if(JudgetLeft(Tree,Tr[Tree].Lc)&&JudgetRight(Tree,Tr[Tree].Rc))
 93         return IsTree(Tr[Tree].Lc)&&IsTree(Tr[Tree].Rc);
 94     else
 95         return 0;
 96 }
 97 
 98 int main()
 99 {
100     int N;
101     scanf("%d", &N);
102     for (int i = 0; i < N; i++)
103     {
104         int K1, K2, Lc, Rc;
105         scanf("%d%d%d%d", &K1, &K2, &Lc, &Rc);
106         Tr[i].K1 = K1;
107         Tr[i].K2 = K2;
108         Tr[i].Lc = Lc;
109         Tr[i].Rc = Rc;
110         if (Lc != -1)
111             Collected[Lc] = 1;
112         if (Rc != -1)
113             Collected[Rc] = 1;
114     }
115     int Tree = FindTree(N);
116     if (IsAVL(Tree) && IsMinHeap(Tree)&&IsTree(Tree))
117         printf("YES");
118     else
119         printf("NO");
120     return 0;
121 }
View Code

 

posted @ 2019-10-10 16:39  57one  阅读(259)  评论(0)  编辑  收藏  举报