PAT-甲级-1135 Is It A Red-Black Tree (30 分)_there is a kind of balanced binary search tree nam-优快云博客

本文链接：https://blog.youkuaiyun.com/valcanoshan/article/details/97965512

该博客介绍了一种平衡二叉搜索树——红黑树，并给出了其五个特性。任务是根据给定的二叉搜索树的先序遍历序列判断是否符合红黑树的条件。输入包含多个测试用例，每个用例包括树的节点数和先序遍历序列。输出应为“是”或“否”，表示给定树是否合法的红黑树。

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1135 Is It A Red-Black Tree (30 分)

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 <vector>
#include <cmath>
using namespace std;
#define rep(i,j,k)    for(int i=j;i<k;i++)
struct node{
    int d;
    node *l,*r;
};
node* build(node *u,int d){
    if(u == NULL){
        u = new node();
        u->d = d;
        u->l = u->r =NULL;
    }else if(abs(d) <= abs(u->d))
        u->l = build(u->l,d);
    else
        u->r = build(u->r,d);
    return u;
}
/* 判断所有的红色节点的孩子是否都为黑色节点 */
bool isChild(node *u){
    if(u == NULL) return true;
    if(u->d < 0){
        if(u->l != NULL && u->l->d < 0) return false;
        if(u->r != NULL && u->r->d < 0) return false;
    }
    return isChild(u->l) && isChild(u->r);
}
/* 计算黑色结点的高度 (同层的黑色结点高度一致) */
int getH(node *u){
    if(u == NULL)    return 0;
    int l = getH(u->l);
    int r = getH(u->r);
    return u->d > 0 ? max(l,r) + 1 : max(l,r);
}
/* 判断所有层的高度是否都平衡（所有黑色节点都在同一层）*/
bool isBlack(node *u){
    if(u == NULL)    return true;
    int l = getH(u->l);
    int r = getH(u->r);
    if(l != r)    return false;
    return isBlack(u->l) && isBlack(u->r);
}
int main(){
    std::ios::sync_with_stdio(false);
    int n,k;
    cin>>n;
    rep(i,0,n){
        cin>>k;
        vector<int> d(k);
        node *u = NULL;
        rep(j,0,k){
            cin>>d[j];
            u = build(u,d[j]);
        }
        if(d[0] < 0 || isChild(u) == false || isBlack(u) == false )
            printf("No\n");
        else printf("Yes\n");
    }
    return 0;
}