PAT甲级 1135. Is It A Red-Black Tree (30)

https://www.patest.cn/contests/pat-a-practise/1135

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

其实这道题我不是很懂，我也不知道怎么通过先序遍历构造平衡二叉树，后来磨了很久做了出来，不过可能方法比较奇怪。

先根据题目所给的先序遍历构造二叉搜索树（其实这样的话就和输入顺序没有关系了），然后逐条检查红黑树的特征，前三条很容易检查，然后traversal1()检查第四条，traversal2()检查第五条。

preordert()输出我构造的二叉树的先序遍历，用来检查是否和输入的先序遍历一样。不过，即使加了这一个功能，得分也没有变化。

第三个测试点不知为什么过不了，丢了一分

#include
#include
using namespace std;

enum color {
	red,
	black
};
struct Node {
	int data;
	color colour;
	Node * left;
	Node * right;
	Node() {
		data = 0;
		colour = black;
		left = NULL;
		right = NULL;
	}
};

void addchild(int num, Node * tree)
{
	int abs = num<0 ? -num : num;
	if (tree->data>abs)
		if (tree->left == NULL)
		{
			Node * newnode = new Node;
			newnode->data = abs;
			newnode->colour = num<0 ? red : black;
			tree->left = newnode;
		}
		else
			addchild(num, tree->left);
	else
	{
		if (tree->right == NULL)
		{
			Node * newnode = new Node;
			newnode->data = abs;
			newnode->colour = num<0 ? red : black;
			tree->right = newnode;
		}
		else
			addchild(num, tree->right);
	}
}

bool traversal1(Node * tree)
{
	bool islegal = true;
	if (tree->colour == red)
		if ((tree->left != NULL&&tree->left->colour == red) || (tree->right != NULL&&tree->right->colour == red))
			return false;
	if (tree->left != NULL)
		islegal &= traversal1(tree->left);
	if (tree->right != NULL)
		islegal &= traversal1(tree->right);
	return islegal;
}

int blacknum = -1;
bool traversal2(Node * tree, int numofblack)
{
	if (tree->colour == black)numofblack++;
	if (tree->left == NULL&&tree->right == NULL)
	{
		if (blacknum == -1)
			blacknum = numofblack;
		else
			if (blacknum != numofblack)
				return false;
	}
	bool islegal = true;
	if (tree->left != NULL)
		islegal &= traversal2(tree->left, numofblack);
	if (tree->right != NULL)
		islegal &= traversal2(tree->right, numofblack);
	return islegal;
}

vector pre;
void preordert(Node * tree)
{
	pre.push_back(tree->colour == black ? tree->data : -tree->data);
	if (tree->left != NULL)
		preordert(tree->left);
	if (tree->right != NULL)
		preordert(tree->right);
}
int main()
{
	int K;
	cin >> K;
	for (int i = 0; i> N;
		int input;
		cin >> input;
		inputs[0] = input;
		if (input <= 0)
			islegal = false;
		treeroot->data = input<0 ? -input : input;
		treeroot->colour = input<0 ? red : black;
		for (int j = 0; j> input;
			inputs[j + 1] = input;
			if (input == 0)
				islegal = false;
			addchild(input, treeroot);
		}
		pre.clear();
		preordert(treeroot);
		for (int j = 0; j