1135 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.

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

思路

注意在建图的时候就有可能不是树

cpp代码

#include<iostream>
#include<unordered_map>
#include<algorithm>
#include<vector>
#include<cstring>
using namespace std;
const int N=50;
unordered_map<int,int> pos,l,r,judge;//负红，正黑
int pre[N],post[N];
bool st=true,st1=false;
vector<int> vec;
int path[N];
int build(int inl,int inr,int prel,int prer){
    int root=pre[prel];
    int k=pos[root];
    if(inl>k||k>inr){
        st1=false;
        return root;
    }
    if(inl<k)l[root]=build(inl,k-1,prel+1,k-1-inl+prel+1);
    if(k<inr)r[root]=build(k+1,inr,k-1-inl+prel+1+1,prer);
    
    return root;
}
void dfs(int u,int deep){
	if(judge[u]==-1){//如果该节点是红的 
		if(l.count(u)){
			if(judge[l[u]]==-1)st=false;
		}
		if(r.count(u)){
			if(judge[r[u]]==-1)st=false;
		}
	}
	
	path[deep]=judge[u];
	
	if(!l.count(u)||!r.count(u)){//注意这里是||而不是&&，因为leaf是空的，因此当没有左孩子或者右孩子时就需要判断
		int num=0;
		for(int i=0;i<=deep;i++)
			if(path[i]==1)num++;
		vec.push_back(num);
	}
	
	if(l.count(u))dfs(l[u],deep+1);
	
	if(r.count(u))dfs(r[u],deep+1);
}
int main(){
    int q;
    cin>>q;
    while(q--){
    	st=true;
    	st1=true;
    	vec.clear();
    	pos.clear();
    	l.clear();
    	r.clear();
        int n;
        cin>>n;
        for(int i=0;i<n;i++){
            int x;cin>>x;
            if(x<0)x=abs(x),judge[x]=-1;
            else judge[x]=1;
            pre[i]=x;
            post[i]=x;
        }
        
        sort(post,post+n);
        
        for(int i=0;i<n;i++)pos[post[i]]=i;
       
        int root=build(0,n-1,0,n-1);
        if(!st1){
             puts("No");
            continue;
        }
        if(judge[root]==-1){
            puts("No");
            continue;
        }
        
        dfs(root,0);
        if(!st){
        	puts("No");
        	continue;
		}
		int t=vec[0];
		bool flag=true;
		for(int i=1;i<vec.size();i++){
			if(vec[i]!=t)flag=false;
		}
		if(!flag){
			puts("No");
			continue;
		}
		puts("Yes");
    }
}