PAT A1135 2019.08.20 【红黑树】

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

思路分析

红黑树作为一种高级数据结构，本科很少涉及，这里也只是提供红黑树的特点，根据这些特点来判断一个树是否是红黑树

此题涉及一些关于二叉树树的基本操作：BST建树、递归求树高 与 遍历的基本操作

此题的判断点有3个 

1、root是否为black

2、red的left、right是否都为black

3、每个node到leaf是否经过相等的black（求树高，只有遇到black才计数）

 

#include<cstdio>
#include<iostream>
#include<cmath>
#define MAX 100
using namespace std;

typedef struct NODE{
	int data;
	struct NODE *left;
	struct NODE *right;
}NODE;

NODE *node[MAX];
int k; 
int n;
int flag=1;

void *insert(NODE *&root,int data)
{
	if(root==NULL)
	{
		NODE *node = new NODE;
		node->left=NULL;
		node->right=NULL;
		node->data=data;
		root=node;
	}
	if(abs(data)<abs(root->data))
	  insert(root->left,data);
	if(abs(data)>abs(root->data))
	  insert(root->right,data);
}

void judgeChild(NODE *root)
{
	if(root==NULL)return;
	if(root->data<0)
	{
		if(root->left!=NULL&&root->left->data<0)flag=0;
		if(root->right!=NULL&&root->right->data<0)flag=0;
    }
	judgeChild(root->left);
	judgeChild(root->right);
}

int getNum(NODE *root)
{
	if(root==NULL)return 0;
	int l=getNum(root->left);
	int r=getNum(root->right);
	return root->data>0?max(l,r)+1:max(l,r);
}
void judgePath(NODE *root)
{
	if(root==NULL)return;
	int l=getNum(root->left);
	int r=getNum(root->right);
	if(l!=r)flag=0;
	judgePath(root->left);
	judgePath(root->right);
}






void preorder(NODE *root)
{
	if(root==NULL)return;
	cout<<root->data<<" ";
	preorder(root->left);
	preorder(root->right);
}

int main()
{
	scanf("%d",&k);
	while(k--)
	{
		flag=1;
		cin>>n;
		NODE *root = new NODE;
		root->left=NULL;root->right=NULL;
		scanf("%d",&root->data);
		if(root->data<0)flag=0;
		for(int i=1;i<n;i++)
		{
			int data;
			cin>>data;
			insert(root,data);
		}
//		preorder(root);
//		cout<<endl;
        judgeChild(root);
        judgePath(root);
        
        if(flag==1)cout<<"Yes"<<endl;
        else cout<<"No"<<endl;
        
	}

}