【题解】【PAT甲】 1135 Is It A Red-Black Tree (30 分)（红黑树）

题目链接

 PTA | 程序设计类实验辅助教学平台

题目描述

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

结尾无空行

Sample Output:

Yes
No
No

结尾无空行

题目大意

 给你一棵平衡二叉树的先序序列，整数代表是黑，负数代表是红，让你判断是不是红黑树

红黑树

1.根节点为黑

2.叶子结点（为NULL）为黑

3.如果一个节点为红结点，它的儿子结点必须为黑节点

4.任意一个结点到左右子树的叶子结点经过的黑节点数目必须相同

解题思路

 一开始写了一个dfs结果跑不起来

可以将一个dfs拆分成两个dfs，一个用来判断条件3，一个判断条件4

条件3没啥好说的，主要是条件4

遍历每一个节点，每个节点的左右子树的黑节点高度必须相等（？也没理解这句话啥意思）

题解

#include<bits/stdc++.h>
using namespace std;
struct Node{
	int val;
	int flag=0;
	int num=0;
	struct Node* left;
	struct Node* right;
	int sub=-1;
};
//0 黑
//1 红 
Node* insert(Node* root,int val,int flag){
	if(root==NULL){
		root=new Node();
		root->left=NULL;
		root->right=NULL;
		root->val=val;
		root->flag=flag;
	}
	else{
		if(val>root->val){
			root->right=insert(root->right,val,flag);
		}
		else{
			root->left=insert(root->left,val,flag);
		}
	}
	return root;
}
vector<Node*> v;
bool dfs1(Node* root,int flag){
	if(root==NULL)	return true;
	if(flag==1&&root->flag==1)	return false;
	return dfs1(root->left,root->flag)&&dfs1(root->right,root->flag);
}
int getNum(Node* root){
	if(root==NULL)	return 1;
	int l=getNum(root->left);
	int r=getNum(root->right);
	return (root->flag==0)?min(l,r)+1:min(l,r);
}
bool dfs2(Node* root){
	if(root==NULL)	return true;
	int left=getNum(root->left);
	int right=getNum(root->right);
	if(left!=right)	return false;
	return dfs2(root->left)&&dfs2(root->right);
}
int main(){
	int t;
	cin>>t;
	while(t--){
		int n;
		scanf("%d",&n);
		int data;
		Node* root;
		int flag;
		for(int i=0;i<n;i++){
			scanf("%d",&data);
			if(data>=0)	flag=0;		//黑 
			else	flag=1;			//红 
			if(i==0){
				root=new Node();
				root->left=NULL;
				root->right=NULL;
				root->val=data;
				root->flag=flag;
			}
			else
				root=insert(root,abs(data),flag);
		}
		if(root->flag==0&&dfs1(root,0)&&dfs2(root))	cout<<"Yes"<<endl;
		else	cout<<"No"<<endl;	 
	}
}