1135 Is It A Red-Black Tree (30 分)

梦想与她 提交于 2020-05-04 03:31:20
1135 Is It A Red-Black Tree (30 分)
 

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.

rbf1.jpg rbf2.jpg rbf3.jpg
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

红黑树也不过是如此嘛,反正都是跟着规则来。
我的output函数基本上解决了所有的判断。

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 int n, m, x;
 4 struct Node
 5 {
 6     int val;
 7     Node *left, *right;
 8 };
 9 set<int> st;
10 Node *insert(Node *root, int val){
11     if(root == NULL){
12         root = new Node();
13         root->val = val;
14         root->left = root->right = NULL;
15     }else{
16         if(abs(root->val) > abs(val)){
17             root->left = insert(root->left, val);
18         }else{
19             root->right = insert(root->right, val);
20         }
21     }
22     return root;
23 }
24 bool output(Node *root, int x){
25     if(root != NULL){
26         int y = root->val<0?0:1;
27         if(root->left != NULL && root->val<0&&root->left->val<0)
28             return false;
29         if(root->right != NULL && root->val<0&&root->right->val<0)
30             return false;
31         if(!output(root->left, x+y)) return false;
32         if(!output(root->right, x+y)) return false;
33     }else{
34         st.insert(x);
35     }
36     return true;
37 }
38 int main(){
39     cin >> n;
40     while(n--){
41         Node *tree = NULL;
42         st.clear();
43         cin >> m;
44         for(int i = 0; i < m; i++){
45             cin >> x;
46             tree = insert(tree, x);
47         }
48         if(tree->val < 0){
49             cout << "No" << endl;
50         }else{
51             bool flag = output(tree, 0);
52             if(flag && st.size() == 1){
53                 cout <<"Yes"<<endl;
54             }else{
55                 cout << "No"<<endl;
56             }
57         }
58     }
59     return 0;
60 }

 






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