PAT(A) 1135. Is It A Red-Black Tree (30)

原题目:

原题链接:https://www.patest.cn/contests/pat-a-practise/1135

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


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

题目大意

本题看似红黑树,实际上不是,算是验证红黑树。给定树的前序遍历,负数代表黑色节点,判断是否为红黑树。

解题报告

由题:判断红黑可以有如下几个条件:

  1. 根必须时黑的。
  2. 所有红节点的孩子都是黑的。
  3. 任意节点到达其子孙叶子节点所有路径上黑色个数是相同的,
    题目中知道先序遍历,又因为他是平衡树,所以中序遍历也是知道的,根据这个建树,然后判断。

代码

1135. Is It A Red-Black Tree (30)
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/*
* Problem: 1135. Is It A Red-Black Tree (30)
* Author: HQ
* Time: 2018-03-13
* State: Done
* Memo: 建树,判断是否为红黑树
*/
#include "iostream"
#include "vector"
#include "algorithm"
using namespace std;

struct Node {
int data = 0;
int blacknum;
struct Node * left = NULL;
struct Node * right = NULL;
};

int K,N;
vector<int> preOrder,inOrder;

int findIn(int x,int s,int l) {
for (int i = 0; i < l; i++) {
if (inOrder[s + i] == x)
return s + i;
}
return -1;
}

struct Node * makeTree(int s1,int s2,int n) {
struct Node * root = NULL;
int pos = findIn(preOrder[s1],s2,n);
if (pos != -1) {
root = new struct Node;
root->data = preOrder[s1];
if (pos - s2 > 0)
root->left = makeTree(s1 + 1, s2, pos - s2);
if (s2 + n - pos - 1 > 0)
root->right = makeTree(s1 + pos - s2 + 1, pos + 1, s2 + n - pos - 1);
}
return root;
}

bool cmp(int x, int y) {
return abs(x) < abs(y);
}

bool dfs(struct Node * root) {
if (root == NULL)
return true;
if (root->data < 0) {
if (root->left != NULL && root->left->data < 0)
return false;
if (root->right != NULL && root->right->data < 0)
return false;
}
bool flag = dfs(root->left) && dfs(root->right);
if (!flag)
return false;
if (root->left == NULL && root->right == NULL)
root->blacknum = (root->data > 0) ? 1 : 0;
int l = (root->left == NULL) ? 0 : root->left->blacknum;
int r = (root->right == NULL) ? 0 : root->right->blacknum;
if (l != r)
return false;
root->blacknum = l + (root->data > 0) ? 1 : 0;
return true;
}

bool check(struct Node * root) {
if (root->data < 0)
return false;
return dfs(root);
}

int main() {
cin >> K;
int x;
for (int i = 0; i < K; i++) {
cin >> N;
preOrder.clear();
inOrder.clear();
for (int j = 0; j < N; j++) {
cin >> x;
preOrder.push_back(x);
inOrder.push_back(x);
}
sort(inOrder.begin(), inOrder.end(), cmp);
struct Node * root = makeTree(0, 0, N);
if (check(root))
cout << "Yes" << endl;
else
cout << "No" << endl;
}
system("pause");
}