A1135 Is It A Red-Black Tree (30 point(s))

红黑树判断

1. 原文

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

2. 解析

• 建树

  create(int x, node* &root)

• 根为红，左右结点为黑

  if(root->data<0). if(root->lchild->data<0) if(root->rchild->data<0)

• 左右黑结点数相同

• 根为黑>0

3. AC代码

#include<cstdio>
#include<algorithm>
using namespace std;
struct node{
	int data;
	node *lchild,*rchild;
};
node* newNode(int x){
	node* root=new node;
	root->data=x;
	root->lchild=root->rchild=NULL;
	return root;
}
void create(int x,node* &root){
	if (root==NULL)
	{
		root=newNode(x);
		return;
	}else if (abs(root->data)>abs(x)){
		create(x,root->lchild);
	}else{
		create(x,root->rchild);
	}
}
bool rRedcBlack(node* root){
	if (root==NULL)
	{
		return true;
	}
	if (root->data<0)
	{
		if (root->lchild!=NULL&&root->lchild->data<0)
		{
			return false;
		}
		if (root->rchild!=NULL&&root->rchild->data<0)
		{
			return false;
		}
	}
	return rRedcBlack(root->lchild)&&rRedcBlack(root->rchild);
}
int blackNum(node* root)
{
	if (root==NULL)
	{
		return 0;
	}
	int leftBlack=blackNum(root->lchild);
	int rightBlack=blackNum(root->rchild);
	return root->data>0?max(leftBlack,rightBlack)+1:max(leftBlack,rightBlack);
}
bool equalBlack(node* root){
	if (root==NULL)
	{
		return true;
	}
	int l=blackNum(root->lchild);
	int r=blackNum(root->rchild);
	if (l!=r)
	{
		return false;
	}
	return equalBlack(root->lchild)&&equalBlack(root->rchild);

}
int main()
{
	int k;
	scanf("%d",&k);
	while(k--){
		int n;
		scanf("%d",&n);
		node* root=NULL;
		int a;
		for (int i = 0; i < n; ++i)
		{
			scanf("%d",&a);
			create(a,root);
		}
		if (rRedcBlack(root)&&equalBlack(root)&&root->data>0)
		{
			printf("Yes\n");
		}else{
			printf("No\n");
		}
	}
	return 0;
}

建树时，判断为abs(root->data)>abs(x)