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

题目大意：

在数据结构中有一种称为红黑树的平衡二叉搜索树。它有以下5个属性：

（1）每个节点都是红色或黑色。
（2）根是黑色的。
（3）每个叶子（NULL）都是黑色的。
（4）如果一个节点是红色的，那么它的两个孩子都是黑色的。
（5）对于每个节点，从节点到后代叶子的所有简单路径包含相同数量的黑色节点。

然后，给一个组平衡二叉树序列，让你分别判断是否为红黑树；

分析：

（1）1性质显然不用判断

（2）2性质的判断就是判断每组序列的第一个元素正负

（3）3号性质不太明白，FIgure1咋与Every leaf (NULL) is black匹配上的？？（这条不用测）

（4）深度遍历创建的BST，遇到红节点且由孩子节点时，进行判断一下即可

（5）先写一个函数统计当前结点到叶子节点的black数量，然后再写一个函数判断当前结点一下左右路径上是否有相同数        量           的黑色结点即可。

虽然给的每组数据是一个完整的BST，但我还是在插入结点时定义的是bool insert（）。最后，使用二叉链表实现上述分析过程 即可。

完整代码：
 

#include<bits/stdc++.h>
using namespace std;
struct node{
	int val;
	node *left,*right;
};
//先序创建BST; 
bool insert(node *&root,int key){
	if(!root){
		root=new node;
		root->val =key;
		root->left =root->right =NULL;
		return true;
	}else if(abs(key)==abs(root->val)) return false;
	 else if(abs(key)<abs(root->val))
	           return insert(root->left,key);
	 else return insert(root->right,key);
}

bool judge1(node *root){
	if(!root) return true;
	if(root->val<0){
		if(root->left!=NULL && root->left->val <0 ) return false;
	    if(root->right!=NULL && root->right->val <0 ) return false;
	}
	return judge1(root->left) && judge1(root->right);
}
//后序遍历获得从当前结点出发到达叶子结点的Black个数
int getblack(node *root){
	if(!root) return 0;
	int l=getblack(root->left);
	int r=getblack(root->right);
	return root->val>0? max(l,r)+1:max(l,r);
}
//检测性质5 
bool judge2(node *root){
	if(!root) return true;
	int l=getblack(root->left);
	int r=getblack(root->right);
	if(l!=r) return false;
	return judge2(root->left)&&judge2(root->right);
} 

int main(){
	int n,k;
	cin>>n;
	for(int i=0;i<n;i++){
		cin>>k;
		vector<int>v;
		v.reserve(k);
		node *root=NULL;
		bool flag=false;
		for(int j=0;j<k;j++){
			cin>>v[j];
			flag=insert(root,v[j]);
		}//分别检验2、4、5三条性质是否否和，flag用于检验是否为BST 
		if(v[0]<0 ||flag==false || judge1(root)==false ||judge2(root)==false)
			cout<<"No\n";
	    else cout<<"Yes\n";
	}
	return 0;
}	

That‘s all !