PAT 1135. Is It A Red-Black Tree

红黑树的性质：

　　(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.

翻译：

　　（1）红黑树节点只能为红或者黑

（2）根节点是黑

（3）null为黑，表示叶子节点（没有child的节点为叶子节点）可以为红为黑。如果条件改为null节点为红，则叶子节点必须为黑。

（4）如果该节点为红，child节点都为黑。

（5）从根节点到叶子节点的每条路径上的总black节点数都相等。

　　吐槽：这次PAT考试由于延考一个小时，又加上临时该题，题目出的真的不咋地，4道题目都是题意不清，全靠不断的猜和提交代码测试，才逐渐摸索出题意。

　　这次题目很简单，第一步根据先序遍历构造数，由于红黑树是BST树，所以已知一个先序就可以很快的构造了。第二步使用dfs来判断是否红黑树就行了。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<vector>
#include<queue>
using namespace std;

int n,a[100];

struct Node
{
    int val;
    int bBlack;
    int lBlackNum;
    int rBlackNum;
    int tBlackNum;
    Node* left;
    Node* right;
    Node()
    {
        left = right = 0;
        lBlackNum = rBlackNum = tBlackNum = 0;
    }
    void setVal(int iVal)
    {
        if(iVal > 0) bBlack = 1;
        else if(iVal < 0)bBlack = 0;
        val = abs(iVal);
    }
};

Node* CreateTree(int l,int r)
{
    if(l > r) return NULL;
    Node* nd = new Node();
    nd -> setVal(a[l]);
    int i = l+1;
    for(;i<=r;++i)
        if(abs(a[i]) > abs(a[l])) break;
    nd -> left = CreateTree(l+1,i-1);
    nd -> right = CreateTree(i,r);
    return nd;
}

void DelTree(Node **nd)
{
    if(*nd == NULL) return;
    DelTree(&(*nd)->left);
    DelTree(&(*nd)->right);
    delete *nd;
    *nd = 0;
}
bool bIsTree = true;

int lastnum = -1;
void dfs(Node* nd,int cnt)
{
    if(!bIsTree) return;
    if(nd == NULL)
    {
        if(lastnum == -1) lastnum = cnt;
        else  if(lastnum != cnt){bIsTree = false;}
        return;
    }
    if(nd->bBlack) ++cnt;
    else
    {
        if(nd->left&&!nd->left->bBlack) bIsTree = false;
        if(nd->right&&!nd->right->bBlack) bIsTree = false;
    }
    dfs(nd->left,cnt);
    dfs(nd->right,cnt);

}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(int i=0;i<n;++i)
            scanf("%d",&a[i]);
        Node* root = CreateTree(0,n-1);
        bIsTree = root->bBlack;
        lastnum = -1;       //初始化会忘
        dfs(root,0);
        if(bIsTree) printf("Yes\n");
        else printf("No\n");
        DelTree(&root);
    }
    return 0;
}