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

坑點：

1. 根可能為紅節點

2. 黑高度不一樣

3. 每個紅節點的兩個子節點必為黑色

#include <iostream>
#include 
 
<string>
#include <map>
#include <set>
#include <list>
#include <vector>
#include <deque>
#include <unordered_set>
#include <algorithm>
#include <unordered_map>
#include <stack>
#include <cstdio>
using namespace std;

int num;
struct treeNode {
    treeNode *left = NULL, *right = NULL;
    int key;
    treeNode(int key):key(key),left(NULL),right(NULL){}
    treeNode(){}
};

void insert(treeNode *root, int key)
{
    auto p = root;
    while (true)
    {
        if (abs(key) > abs(p->key))
        {
            if (p->right == NULL)
            {
                p->right = new treeNode(key);
                break;
            }
            p = p->right;
        }
        else
        {
            if (p->left == NULL)
            {
                p->left = new treeNode(key);
                break;
            }
            p = p->left;
        }
    }
}
void dfs(treeNode *tn,int cnt,bool &success)
{
    if (tn == NULL)
    {
        if (num == -1)
            num = cnt;
        else if (num != cnt)    //判斷從根節點到所有葉節點的黑高度
            success = false;
        return;
    }
    if (tn->key > 0)
        cnt++;    //黑高度+1
    else{
        if (tn->left != NULL&&tn->left->key < 0)    //判斷紅節點的兒子是否還是紅節點
            success = false;
        if (tn->right != NULL&&tn->right->key < 0)
            success = false;
    }
    dfs(tn->left, cnt, success);
    dfs(tn->right, cnt, success);
}
int main()
{
    int num_case;
    cin >> num_case;
    
    while (num_case--)
    {
        treeNode* root = new treeNode;
        int num_nodes;
        bool success = true;
        
        cin >> num_nodes;
        int nodeKey;
        cin >> nodeKey;
        root->key = nodeKey;
        num_nodes--;
        while (num_nodes--)
        {
            cin >> nodeKey;
            insert(root, nodeKey);
        }
        num = -1;
        dfs(root,0,success);
        if (root->key < 0)    //根為紅節點
            success = false;
        string s = success ? "Yes" : "No";
        cout << s << endl;
    }
    
    return 0;
}

【PAT1135】Is It A Red-Black Tree（30）