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

 

#include<bits/stdc++.h>
using namespace std;
const int maxn=1010;
#define  inf  0x3fffffff
vector<int> vis;
typedef struct node{
    int data;
    struct node *lchild;
    struct node *rchild;
}node,*pnode;
void build(pnode &root,int temp){
    if(root==NULL){
        root=new node;
        root->data=temp;
        root->lchild=root->rchild=NULL;
        return ;
    }
    else if(abs(temp)<=abs(root->data)){
        build(root->lchild,temp);
    }
    else{
        build(root->rchild,temp);
    }

}
bool judge1(pnode root){
    if(root==NULL){
        return true;
    }
    if(root->data<0){
        if(root->lchild!=NULL&&root->lchild->data<0){
            return false;
        }
        if(root->rchild!=NULL&&root->rchild->data<0){
            return false;
        }
    }
    return judge1(root->lchild)&&judge1(root->rchild);
}
int getNum(pnode root){
    if(root==NULL){
        return 0;
    }
    int l=getNum(root->lchild);
    int r=getNum(root->rchild);
    return root->data>0?max(l,r)+1:max(l,r);
}
bool judge2(pnode root){
    if(root==NULL){
        return true;
    }
    int l=getNum(root->lchild);
    int r=getNum(root->rchild);
    if(l!=r){
        return false;
    }
    return judge2(root->lchild)&&judge2(root->rchild);
}
int main(){
    int n,m;
    scanf("%d",&n);
    while(n--){
        pnode root=NULL;
        scanf("%d",&m);
        vis.resize(n);
        for(int i=0;i<m;i++){
            scanf("%d",&vis[i]);
            build(root,vis[i]);
        }
        if(vis[0]<0||judge1(root)==false||judge2(root)==false){
            printf("No
");
        }
        else{
            printf("Yes
");
        }
    }
    return 0;
}