PAT 1135 Is It A Red-Black Tree

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 ( $\leq30) which is the total number of cases. For each case, the first line gives a positive integer N (\leq30), the total number of nodes in the binary tree. The second line gives the preorder traversal sequence of the tree. While all the keys in a tree are positive integers, we use negative signs to represent red nodes. All the numbers in a line are separated by a space. The sample input cases correspond to the trees shown in Figure 1, 2 and 3.$

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 }