[数据结构与算法]*PAT甲级1135 Is It A Red-Black Tree |判断红黑树

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.



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

要判断所给的BST是否符合下面性质：
（1）每个节点或者是黑色，或者是红色。
（2）根节点是黑色。
（3）每个叶子节点（NIL）是黑色。 [注意：这里叶子节点，是指为空(NIL或NULL)的叶子节点！]
（4）如果一个节点是红色的，则它的子节点必须是黑色的(从每个叶子到根的所有路径上不能有两个连续的红色节点) 。
（5）从任一节点到其每个叶子节点（NULL）的所有简单路径都包含相同数目的黑色节点。

思路

不太会做，只会建树和判断性质4，性质5不会判断，主要参考1135 Is It A Red-Black Tree(判断是否是红黑树)

#include <iostream>
#include <cstdlib>
#include <vector>
#include <algorithm>
using namespace std;
struct node
{
	int val;
	node* left;
	node* right;
	node(int v) : val(v), left(NULL), right(NULL) {}
};
int pre[31], in[31];
node* creat(int preL, int preR, int inL, int inR);
bool judge4(node* root); //判断性质4
bool judge5(node* root); //判断性质5
int cntBlack(node* root);
int main()
{
	int K;
	cin >> K;
	for (int i = 0; i < K; i++)
	{
		int N;
		cin >> N;
		for (int j = 0; j < N; j++)
		{
			cin >> pre[j];
			in[j] = abs(pre[j]);
		}
		sort(in, in + N - 1);
		node* root = creat(0, N - 1, 0, N - 1);
		if (pre[0] > 0 && judge4(root) && judge5(root))
			cout << "Yes" << endl;
		else
			cout << "No" << endl;
	}
	system("pause");
	return 0;
}
node* creat(int preL, int preR, int inL, int inR)
{
	if (preL > preR)
		return NULL;
	node* root = new node(pre[preL]);
	int k;
	for (k = inL; k <= inR; k++)
		if (in[k] == abs(pre[preL]))
			break;
	int numLeft = k - inL;
	root->left = creat(preL + 1, preL + numLeft, inL, k - 1);
	root->right = creat(preL + numLeft + 1, preR, k + 1, inR);
	return root;
}
bool judge4(node* root)
{
	if (!root)
		return true;
	if (root->val < 0)
	{
		//左孩子或右孩子是红的说明违反性质4
		if (root->left != NULL && root->left->val < 0) 
			return false;
		if (root->right != NULL && root->right->val < 0)
			return false;
	}
	return judge4(root->left) && judge4(root->right);
}
bool judge5(node* root)
{
	if (!root)
		return true;
	int leftBlackNum = cntBlack(root->left);
	int rightBlackNum = cntBlack(root->right);
	if (leftBlackNum != rightBlackNum)
		return false;
	return judge5(root->left) && judge5(root->right);
}
int cntBlack(node* root)
{
	if (!root)
		return 0;
	int left = cntBlack(root->left);
	int right = cntBlack(root->right);
	if (root->val > 0) //只数黑色
		return max(left, right) + 1;
	else
		return max(left, right);
}