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

 

題解: 

 

#include <iostream>
#include <algorithm>
#include <unordered_map>

using namespace std;

const int N = 40;

int pre[N], in[N];
unordered_map<int, int> pos;
bool ans;

int build(int il, int ir, int pl, int pr, int& sum)
{
    int root = pre[pl];
    int k = pos[abs(root)];

    if (k < il || k > ir)
    {
        ans = false;
        return 0;
    }

    int left = 0, right = 0, ls = 0, rs = 0;
    if (il < k) left = build(il, k - 1, pl + 1, pl + 1 + k - 1 - il, ls);
    if (k < ir) right = build(k + 1, ir, pl + 1 + k - 1 - il + 1, pr, rs);

    if (ls != rs) ans = false;
    sum = ls;
    if (root < 0)
    {
        if (left < 0 || right < 0) ans = false;
    }
    else sum ++ ;

    return root;
}

int main()
{
    int T;
    cin >> T;
    while (T -- )
    {
        int n;
        cin >> n;
        for (int i = 0; i < n; i ++ )
        {
            cin >> pre[i];
            in[i] = abs(pre[i]);
        }

        sort(in, in + n);

        pos.clear();
        for (int i = 0; i < n; i ++ ) pos[in[i]] = i;

        ans = true;
        int sum;
        int root = build(0, n - 1, 0, n - 1, sum);

        if (root < 0) ans = false;
        if (ans) puts("Yes");
        else puts("No");
    }

    return 0;
}