A1135. 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 (≤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

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<queue>
using namespace std;
bool cmp(int a, int b){
    return abs(a) < abs(b);
}
int K, N;
typedef struct NODE{
    struct NODE *lchild, *rchild;
    int data;
}node;
void insert(node* &root, int data){
    if(root == NULL){
        root = new node;
        root->lchild = NULL;
        root->rchild = NULL;
        root->data = data;
        return;
    }
    if(abs(data) < abs(root->data))
        insert(root->lchild, data);
    else insert(root->rchild, data);
}
int cnt, isEqu;
void preOrder(node* root, int dp){
    if(root == NULL){
        dp++;
        if(cnt == -1){
            cnt = dp;
        }else{
            if(cnt != dp)
                isEqu = 0;
        }
        return;
    }
    if(root->data > 0)
        dp++;
    preOrder(root->lchild, dp);
    preOrder(root->rchild, dp);
}
int exam(node* root){
    if(root->data < 0) //负数为红
        return 0;
    queue<node*> Q;
    Q.push(root);
    int tag = 1;
    while(Q.empty() == false){
        node* temp = Q.front();
        if(temp->data < 0){
            if(temp->lchild != NULL && temp->lchild->data < 0 || temp->rchild != NULL && temp->rchild->data < 0){
                tag = 0;
                break;
            }
        }
        Q.pop();
        cnt = -1, isEqu = 1;
        preOrder(temp, 0);
        if(isEqu == 0){
            tag = 0;
            break;
        }
        if(temp->lchild != NULL)
            Q.push(temp->lchild);
        if(temp->rchild != NULL)
            Q.push(temp->rchild);
    }
    return tag;
}
int main(){
    scanf("%d", &K);
    for(int i = 0; i < K; i++){
        scanf("%d", &N);
        node* root = NULL;
        for(int j = 0; j < N; j++){
            int temp;
            scanf("%d", &temp);
            insert(root, temp);
        }
        if(root == NULL)
            printf("Yes
");
        else if(exam(root) == 1)
            printf("Yes
");
        else printf("No
");
    }
    cin >> N;
    return 0;
}

总结：

1、题意：给出一个平衡二叉搜索树的前序序列，给出红黑树的定义，检验该平衡二叉搜索树是否是红黑树。

2、给出了平衡二叉搜索树的前序序列，就可以仅仅根据前序序列建立原树，再按部就班进行检验。检验可以分别针对红黑树的要求逐条检验，首先看根。然后按照层序的顺序，对每一个节点做如下检验：1）若它是红的，检验它的左右孩子。 2）用DFS，遍历从该节点开始到叶节点（空节点）的所有路径，统计每个路径分别的黑节点总数。

3、关于建立原树，有两种办法。一是，由于搜索树的中序是从小到大的有序序列，可以先将所有节点排序得到中序序列。再按照已知前序和中序的方法，建立二叉树。二是，由于有序二叉树的先序序列的意义：根在前子树在后，且小于根的节点在左，大于的在右。所以可以直接把先序序列当作有序二叉树的插入的顺序，按顺序插入节点，得到原树。注意已知序列是有序二叉树的先序，则可以把它当作插入顺序。但已知插入顺序，这个插入顺序却不一定是先序。