1135.Is It A Red-Black Tree

题目描述

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.

Figure1

Figure2

Figure3

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

考点

1.先序遍历；
2.红黑树；
3.二叉搜索树。

思路

1.递归实现
红黑树的任意子树中，根结点到叶子结点所经过的黑结点数也相等，因此可以递归实现。
2.具体实现
a.判断根结点是否为黑色；
b.如果当前子树的根结点为红色，则判断其左右孩子结点是否均为黑色。根据先序遍历可以轻松定位其左右孩子。假设当前结点为先序遍历数组的第i个位置，那么其左孩子在i+1个位置，右孩子在第一个值比当前值大的位置；
c.如果当前根结点的左右子树经过的黑色结点的个数num不一致则返回0；否则，若当前根结点为黑色，则返回num+1，红色则返回num即可；
d.叶子结点根据当前颜色返回值。

代码

#include 
#include 
using namespace std;
vector pre;
int judge(int l, int r) {
    //root is red
    if (l == 0 && pre[l] < 0)return 0;
    //leaf
    if (l == r) {
        if (pre[l] > 0) return 1;
        else return 0;
    }
    if (l > r) return 0;
    int i = l+1;
    while (abs(pre[i]) < abs(pre[l])) i++; 
    //node is red and its children not all black
    if ((pre[l] < 0) && (pre[l+1] < 0 || pre[i] < 0)) return 0;
    int lnum = judge(l + 1, i - 1);
    int rnum = judge(i, r);
    if (lnum == rnum) return lnum + (pre[l] > 0 ? 1 : 0);
    else return 0;
}
int main() {
    int k; cin >> k;
    for (int i = 0; i < k; i++) {
        int n; cin >> n;
        pre.resize(n);
        for (int j = 0; j < n; j++)  cin >> pre[j];
        if (judge(0, n-1)) cout << "Yes" << endl;
        else cout << "No" << endl;
    }
    return 0;
}