1135 Is It A Red-Black Tree 1135 Is It A Red-Black Tree（30 分）

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

题意：

红黑树是平衡二叉搜索树的一种，满足以下条件

1.每个节点非黑即红

2.根结点是黑色的

3.每个叶（空）节点是黑色的【这个条件我至今不知道什么用.....不是很明白】

4.如果一个结点是红色，则它的孩子结点一定是黑的【即没有两个连续的红色结点】

5.满足每个左右子树黑色结点数相同

思路：

1.建平衡二叉树

2.判断是否有连续的红色结点

3.判断左右子树黑色结点数是否相同

代码：

#include <bits/stdc++.h>
using namespace std;
typedef struct node{
    int v;
    struct node *left,*right;
}Node;
int a[33];
int getBlackNum(Node*tre)
{
    if(tre==NULL)return 0;
    int l=getBlackNum(tre->left);
    int r=getBlackNum(tre->right);
    if(tre->v>0)
    {
        return max(l,r)+1;
    }
    else{
        return max(l,r);
    }
}
Node*Insert(Node*tre,int val)
{
    if(tre==NULL)
    {
        tre=new Node();
        tre->v=val;
        return tre;
    }
    if(abs(val)<=abs(tre->v))//插在左子树
    {
        tre->left=Insert(tre->left,val);
    }
    else{
        tre->right=Insert(tre->right,val);
    }
    return tre;
}
bool judgeRed(Node*tre)
{
    if(tre==NULL)return true;
    //Red Red
    if(tre->v<0){
    if(tre->left!=NULL&&tre->left->v<0)return false;
    if(tre->right!=NULL&&tre->right->v<0)return false;
    }
    return judgeRed(tre->left)&&judgeRed(tre->right);
}
bool judgeNum(Node*tre)
{
    if(tre==NULL)return true;
    int l=getBlackNum(tre->left);
    int r=getBlackNum(tre->right);
    if(l!=r)return false;
    return judgeNum(tre->left)&&judgeNum(tre->right);
}
int main()
{
    int t;scanf("%d",&t);
    while(t--)
    {
        int n;scanf("%d",&n);
        Node *tre=NULL;
        for(int i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
            tre=Insert(tre,a[i]);
        }
        if(a[0]>0&&judgeNum(tre)&&judgeRed(tre))printf("Yes\n");
        else printf("No\n");
    }
}