PAT 1135 Is It A Red-Black Tree

Posted 2021-01-01 mr-stn

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

题目大意: 根据前序遍历构建二叉树， 判断该二叉树数是否为红黑树
注意点：红黑树并不是AVL
根据题目中的五个条件就能判断该树是否是红黑树

 1 #include<iostream>
 2 #include<vector>
 3 #include<algorithm>
 4 #include<queue>
 5 using namespace std;
 6 #define max(a,b)((a>b)?(a):(b))
 7 #define abs(a)((a<0)?(0-a):(a))
 8 
 9 struct Node{
10   int val;
11   bool isRed;
12   Node *left, *right;
13 };
14 
15 Node* insert(Node* root, int val){
16   if(root==NULL){
17     root = new Node();
18       root->val = val;
19       root->isRed = val<0 ? true : false;
20       root->left = root->right = NULL;
21   }
22   else if(abs(val)<abs(root->val)) root->left = insert(root->left, val);
23   else root->right = insert(root->right, val);
24   return root;
25 }
26 //判断条件3，4
27 bool isRedBlack(Node* root){
28     if(root==NULL) return true;
29     if(root->isRed && ((root->left!=NULL && root->left->isRed)||(root->right!=NULL && root->right->isRed))) return false;
30     return isRedBlack(root->left) && isRedBlack(root->right);
31 }
32 //计算路径上黑节点的个数
33 int pathLength(Node* root){
34     if(root==NULL) return 0;
35     int l=pathLength(root->left), r=pathLength(root->right);
36     return root->isRed ? max(l, r) : max(l, r)+1;
37 }
38 //判断条件5
39 bool pathEqual(Node* root){
40     if(root==NULL) return true;
41     int l=pathLength(root->left), r=pathLength(root->right);
42     if(l!=r) return false;
43     return pathEqual(root->left) && pathEqual(root->right);
44 }
45 
46 int main(){
47   int n, i, j;
48   scanf("%d", &n);
49   for(i=0; i<n; i++){
50     int cnt;
51     scanf("%d", &cnt);
52     vector<int> v(cnt);
53     Node *root = NULL;
54     for(j=0; j<cnt; j++){
55       scanf("%d", &v[j]);
56       root = insert(root, v[j]);
57     }
58     if(!root->isRed  && isRedBlack(root) && pathEqual(root)) printf("Yes
");
59     else printf("No
");
60   }
61   return 0;
62 }