PAT 1135 Is It A Red-Black Tree

原题链接：1135 Is It A Red-Black Tree (30分)
关键字：红黑树、BST树建树
参考的浒鱼鱼的博客：17年春季第四题 PAT甲级 1135 Is It A Red-Black Tree (30)

There is a kind of balanced binary search tree named red-black tree in the data structure. It has the following 5 properties:

• Every node is either red or black.
• The root is black.
• Every leaf (NULL) is black.
• If a node is red, then both its children are black.
• 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： 给出一棵二叉搜索树的前序遍历，请你判断它是否是红黑树。
分析：
要判断是否满足红黑树五个定义：

1. Every node is either red or black. （不用判断）
2. The root is black.（根结点的值大于0）
3. Every leaf (NULL) is black.（不用判断，但是可以用在4的判断里）
4. If a node is red, then both its children are black.（需要判断）
5. For each node, all simple paths from the node to descendant leaves contain the same number of black nodes.（需要判断）

注意： 红黑树是一种二叉搜索树（BST）所以中序唯一确定，那给定前序就能确定一棵树了。

红黑树属于平衡二叉树。说它不严格是因为它不是严格控制左、右子树高度或节点数之差小于等于1。也就是说**红黑树并非严格控制左右孩子高度相差在1以内。**但红黑树高度依然是平均log(n),且最坏情况高度不会超过2log(n),这有数学证明。所以它算平衡树,只是不严格。不过严格与否并不影响数据结构的复杂度。

步骤：

1. 根据先序建立一棵树，用链表表示（常规的BST的建树方法）
2. 判断根结点（题目所给先序的第一个点即根结点）是否是黑色
3. 根据建立的树，从根结点开始遍历，如果当前结点是红色，判断它的孩子节点是否为黑色，递归返回结果
4. 从根节点开始，递归遍历，检查每个结点的左子树的高度和右子树的高度（这里的高度指黑色结点的个数），比较左右孩子高度是否相等，递归返回结果

代码：

#include 
#include 
#include 
using namespace std;

const int maxn = 40;
int k, n;   //k询问数 n结点数
vector<int> arr;

struct node{
    int value;
    struct node *l, *r;
};

//建树
struct node* build(node *root, int v){
    if(root == NULL) {
        root = new node();
        root->value = v;
        root->l = root->r = NULL;
    }
    else if(abs(v) <= abs(root->value)) root->l = build(root->l, v);
    else root->r = build(root->r, v);
    return root;
}

//从根结点开始遍历，如果当前结点是红色，判断它的孩子节点是否为黑色，递归返回结果
bool judge1(node *root) {
    if (root == NULL) return true;
    if (root->value < 0) {
        if (root->l != NULL && root->l->value < 0) return false;
        if (root->r != NULL && root->r->value < 0) return false;
    }
    return judge1(root->l) && judge1(root->r);
}
//求出节点的高度（这里的高度指黑色结点的个数）
int getNum(node *root) {
    if (root == NULL) return 0;
    int l = getNum(root->l);
    int r = getNum(root->r);
    return root->value > 0 ? max(l, r) + 1 : max(l, r);
}
//检查每个结点的左子树的高度和右子树的高度（这里的高度指黑色结点的个数），比较左右孩子高度是否相等，递归返回结果
bool judge2(node *root) {
    if (root == NULL) return true;
    int l = getNum(root->l);
    int r = getNum(root->r);
    if(l != r) return false;
    return judge2(root->l) && judge2(root->r);
}

int main(){
    cin >> k;
    for(int i = 0; i < k; i ++ ){
        cin >> n;
        arr.resize(n);
        node *root = NULL;
        for(int i = 0; i < n; i ++ ){
            cin >> arr[i];
            root = build(root, arr[i]);
        }
        if(arr[0] < 0 || judge1(root) == false || judge2(root) == false) puts("No");
        else puts("Yes");

    }
    
    return 0;
}