1135 Is It A Red-Black Tree (30 分) 判斷紅黑樹

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

題意：給定一棵樹的前序遍歷，負值爲紅色，判斷是否爲紅黑樹。

紅黑樹的要求：1根節點爲黑色；2每個紅色結點的子節點爲黑色；3點到葉子結點的黑色數相同

思路：一開始以爲紅黑樹必是AVL，傻乎乎調整了半年，然而並不需要... 她只是一棵BST...

首先常規BST插入結點操作，然後以此判斷條件123

#include<stdio.h>
#include<math.h>
#include<string>
#include<vector>
#include<string.h>
#include<iostream>
#include<sstream>
#include<algorithm>
using namespace std;
int color[100005];
struct node
{
	int data;
	node *l,*r;
};
node *insert(node *root,int data)
{
	if(root==NULL)
	{
		root=new node();
		root->data=data;
		root->l=root->r=NULL;
	}
	else if(data<root->data)
	root->l=insert(root->l,data);
	else
	root->r=insert(root->r,data);
	return root;
}
bool dfs(node *root)
{
	if(root==NULL) return true; 
	if(color[root->data])
	{
		if(root->l!=NULL&&color[root->l->data])
		return false;
		if(root->r!=NULL&&color[root->r->data])
		return false;
	}
	return dfs(root->l)&&dfs(root->r);
}
int getNum(node *root) 
{
	if (root==NULL) return 0;
	int left=getNum(root->l);
	int right=getNum(root->r);
	return color[root->data]==0 ? max(left,right)+1 : max(left,right);
}
bool judge(node *root) 
{
	if(root==NULL) return true; 
	int left=getNum(root->l);
	int right=getNum(root->r);
	if(left!=right) return false;
	return judge(root->l)&&judge(root->r);
}
int main()
{
	int t,n,x,i;
	scanf("%d",&t);
	while(t--)
	{
		memset(color,0,sizeof color);
		node *root=NULL;
		scanf("%d",&n);
		for(i=0;i<n;i++)
		{
			scanf("%d",&x);
			if(x<0) x=-x,color[x]=1;
			root=insert(root,x);
		}		
		if(color[root->data]==1)//判斷根節點爲黑
		printf("No\n");
		else
		{
			if(dfs(root)&&judge(root))
			printf("Yes\n");
			else
			printf("No\n");
		}
	}	
}