PAT_A1135#Is It A Red-Black Tree

Source:

PAT A1135 Is It A Red-Black Tree (30 分)

Description:

There is a kind of balanced binary search tree named red-black tree in the data structure. It has the following 5 properties:

• (1) Every node is either red or black.
• (2) The root is black.
• (3) Every leaf (NULL) is black.
• (4) If a node is red, then both its children are black.
• (5) For each node, all simple paths from the node to descendant leaves contain the same number of black nodes.

For example, the tree in Figure 1 is a red-black tree, while the ones in Figure 2 and 3 are not.

Figure 1	Figure 2	Figure 3

For each given binary search tree, you are supposed to tell if it is a legal red-black tree.

Input Specification:

Each input file contains several test cases. The first line gives a positive integer K (≤30) which is the total number of cases. For each case, the first line gives a positive integer N (≤30), the total number of nodes in the binary tree. The second line gives the preorder traversal sequence of the tree. While all the keys in a tree are positive integers, we use negative signs to represent red nodes. All the numbers in a line are separated by a space. The sample input cases correspond to the trees shown in Figure 1, 2 and 3.

Output Specification:

For each test case, print in a line "Yes" if the given tree is a red-black tree, or "No" if not.

Sample Input:

3
9
7 -2 1 5 -4 -11 8 14 -15
9
11 -2 1 -7 5 -4 8 14 -15
8
10 -7 5 -6 8 15 -11 17

Sample Output:

Yes
No
No

Keys:

• 二叉查找树（Binary Search Tree）

Attention:

• 应该注意到，红黑树（ balanced binary search tree）并不是AVL树（self-balancing binary search tree），英文题描述题这里很容易产生误解；
• tips1：树的结点值置为先序序列的下标，而非实际的键值；
• tips2：pre[0]设为根结点的根结点，且为红色。这样从根结点遍历时，不必特判性质2，由性质4可以推出性质2；
• tips3：ans=0时剪枝，减少运行时间；

Code:

 1 /*
 2 Data: 2019-05-28 19:30:11
 3 Problem: PAT_A1135#Is It A Red-Black Tree
 4 AC: 01:01:28
 5 
 6 题目大意：
 7 红黑树具有如下性质的平衡二叉树：
 8 1.结点非红即黑
 9 2.根结点为黑色
10 3.叶子结点（空结点）为黑色
11 4.红色结点的孩子均为黑色
12 5.任意结点到叶子结点构成的简单路径所含的黑色结点数目相同
13 现给定一颗二叉树，判定其是否为红黑树
14 输入：
15 第一行给出，测试数K<=20
16 第二行给出，结点总数N<=30
17 第三行给出，先序遍历，键值均正，负数表示红色结点
18 */
19 
20 #include<cstdio>
21 #include<cmath>
22 using namespace std;
23 const int M=1e2;
24 int pre[M],ans;
25 struct node
26 {
27     int index;
28     node *lchild,*rchild;
29 };
30 
31 node* Create(node *root, int index)
32 {
33     if(!root)
34     {
35         root = new node;
36         root->index=index;
37         root->lchild=root->rchild=NULL;
38     }
39     else if(abs(pre[root->index]) > abs(pre[index]))
40         root->lchild = Create(root->lchild,index);
41     else
42         root->rchild = Create(root->rchild,index);
43     return root;
44 }
45 
46 int Travel(node *root, int fa)
47 {
48     if(!ans || !root)
49         return 0;
50     int l = Travel(root->lchild,root->index);
51     int r = Travel(root->rchild,root->index);
52     if(l!=r || (pre[fa]<0&&pre[root->index]<0)){
53         ans=0;return 0;
54     }
55     if(pre[root->index]>0)
56         l++;
57     return l;
58 }
59 
60 int main()
61 {
62 #ifdef    ONLINE_JUDGE
63 #else
64     freopen("Test.txt", "r", stdin);
65 #endif
66 
67     int k,n;
68     scanf("%d", &k);
69     while(k--)
70     {
71         scanf("%d", &n);
72         node *root=NULL;
73         for(int i=1; i<=n; i++){
74             scanf("%d", &pre[i]);
75             root=Create(root, i);
76         }
77         ans=1;pre[0]=-1;
78         Travel(root,0);
79         if(ans)
80             printf("Yes\n");
81         else
82             printf("No\n");
83     }
84 
85     return 0;
86 }