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
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 }