PAT 甲级 1135 Is It A Red-Black Tree

1135 Is It A Red-Black Tree （30 point(s)）

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

经验总结：

首先，没AC的请注意，红黑树不是AVL树！不需要进行平衡判定！这一点我也是看了柳神的解释才知道，坑爹的百度百科木有明确的说明= =，不用平衡判定，这一题就很简单了，红黑树五个条件，但是其中，结点非黑即红，以及所有叶结点（空结点）为黑这两个条件无需判定，只需要判定根结点为黑结点，所有的红结点的孩子必须为黑结点，以及所有结点往下访问至叶结点路径中所经过的黑结点数目相同三个条件~

AC代码

#include <cstdio>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;
int n,m,k,num;
bool flag;
struct node
{
	int data;
	node *lchild,*rchild;
};
void insert(node *&root,int data)
{
	if(root==NULL)
	{
		root=new node();
		root->lchild=root->rchild=NULL;
		root->data=data;
		return ;
	}
	if(abs(data)<=abs(root->data))
		insert(root->lchild,data);
	else
		insert(root->rchild,data);
}
void DFS(node *root,int cnt)
{
	if(root==NULL)
	{
		if(flag==true)
			if(num==-1)
				num=cnt;
			else if(num!=cnt)
				flag=false;
		return ;
	}
	if(root->data>=0)
		++cnt;
	DFS(root->lchild,cnt);
	DFS(root->rchild,cnt);
}
bool BFS(node *root)
{
	queue<node *> q;
	if(root->data<0)
		return false;
	q.push(root);
	while(q.size())
	{
		node *x=q.front();
		flag=true;
		num=-1;
		DFS(x,0);
		if(flag==false)
			return flag;
		q.pop();
		if(x->data<0&&(x->lchild!=NULL&&x->lchild->data<0||x->rchild!=NULL&&x->rchild->data<0))
			return false;
		if(x->lchild!=NULL)
			q.push(x->lchild);
		if(x->rchild!=NULL)
			q.push(x->rchild);
	}
	return true;
}

int main()
{
	scanf("%d",&k);
	for(int i=0;i<k;++i)
	{
		node* root=NULL;
		scanf("%d",&n);
		for(int j=0;j<n;++j)
		{
			scanf("%d",&m);
			insert(root,m);
		}
		printf("%s\n",BFS(root)?"Yes":"No");
	}
	return 0;
}