1.题目
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
No2.题目分析
1.红黑树：
- 性质1：每个节点要么是黑色，要么是红色。
- 性质2：根节点是黑色。
- 性质3：每个叶子节点（NIL）是黑色。
- 性质4：每个红色结点的两个子结点一定都是黑色。
- 性质5：任意一结点到每个叶子结点的路径都包含数量相同的黑结点。
参考：https://www.jianshu.com/p/e136ec79235c
2.建树
红黑树的中序遍历是从小到大的（建树用了两种小小不同的形式）
3.判断
计算从底部到根的黑节点的个数（未将为空的叶子节点算入，但是不影响结果），同时看一个红节点的两个子节点，要不两个都为空，要不两个都是黑节点，否则为NO
3.代码
#include<iostream>
#include<vector>
#include<cmath>
#include<algorithm>
using namespace std;
typedef struct node* tree;
struct node
{
int data;
tree left;
tree right;
};
tree creat(tree t,int root, int ins,int ine,vector<int>pre,vector<int>in,vector<int>mark)
{
if (ins>ine)return NULL;
if (t == NULL)t = (tree)malloc(sizeof(struct node));
t->data = pre[root]*mark[root];
t->left = t->right = NULL;
int i = ins;
while (in[i] != pre[root])i++;
if(i<=ine)
{t->left=creat(t->left, root + 1, ins,i-1, pre, in, mark);
t->right = creat(t->right, root + i + 1-ins, i + 1,ine, pre,in,mark);
}
return t;
}
// tree creat(tree t, int pres,int pree, int ins, int ine, vector<int>pre, vector<int>in, vector<int>mark)
// {
// if (pres>pree)return NULL;
// if (t == NULL)t = (tree)malloc(sizeof(struct node));
// t->data = pre[pres] * mark[pres];
// t->left = t->right = NULL;
// int i = ins;
// while (i<ine&&in[i] != pre[pres])i++;
// t->left = creat(t->left, pres + 1,pres+i-ins, ins, i - 1, pre, in, mark);
// t->right = creat(t->right, pres + i + 1 - ins,pree, i + 1, ine, pre, in, mark);
// return t;
// }
bool ok = true;
int judge(tree t)
{
int l, r;
if (t == NULL)return 0;
l = judge(t->left);
r = judge(t->right);
if (l != r) { ok = false; return l; }
if (t->data < 0)
{
if ((t->left == NULL&&t->right == NULL) || (t->left != NULL&&t->right != NULL&&t->left->data>0 && t->right->data > 0));
else ok = false;
}
if(t->data>0)
return l+1;
else return l;
}
int main()
{
int k,n,temp;
scanf("%d", &k);
for (int i = 0; i < k; i++)
{
ok = true;
scanf("%d", &n);
vector<int>pre,in;
vector<int>mark(n, 1);
for (int j = 0; j < n; j++)
{
scanf("%d", &temp);
if (temp < 0)mark[j] = -1;
pre.push_back(abs(temp));
in.push_back(abs(temp));
}
sort(in.begin(), in.end());
tree t = NULL;
t = creat(t,0, 0,n-1, pre, in, mark);
//t = creat(t, 0,n-1, 0, n - 1, pre, in, mark);
if (t->data < 0) { printf("No\n"); continue; }
judge(t);
if (ok)printf("Yes\n");
else printf("No\n");
}
}