• 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.

    rbf1.jpgrbf2.jpgrbf3.jpg
    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

    代码:
     1 #include <bits/stdc++.h>
     2 using namespace std;
     3 struct node
     4 {
     5     int val;
     6     node* l, *r;
     7 };
     8 int v[40];
     9 node* root;
    10 
    11 void build(node *now, int val)
    12 {
    13     int v = now->val;
    14     if (abs(val) < abs(v))
    15     {
    16         if (now->l == NULL)
    17             now->l = new node(), now->l->val = val;
    18         else build(now->l, val);
    19     }
    20     else
    21     {
    22         if (now->r == NULL)
    23             now->r = new node(), now->r->val = val;
    24         else build(now->r, val);
    25     }
    26 }
    27 
    28 bool judge(node* x)
    29 {
    30     if (!x) return true;
    31     if (x->val < 0)
    32         if (x->l && x->l->val < 0 || x->r && x->r->val < 0) return false;
    33     return judge(x->l) && judge(x->r);
    34 }
    35 
    36 int cal(node* x)
    37 {
    38     if (!x) return 0;
    39     int numl = cal(x->l), numr = cal(x->r);
    40     if (numl == numr && numl != -1)
    41         return x->val > 0 ? numl + 1 : numl;
    42     return -1;
    43 }
    44 int main()
    45 {
    46     int t; cin >> t;
    47     while (t--)
    48     {
    49         int n; cin >> n;
    50         for (int i = 1; i <= n; i++)
    51             cin >> v[i];
    52         root = new node();
    53         root->val = v[1], root->l = root->r = NULL;
    54         for (int i = 2; i <= n; i++)
    55             build(root, v[i]);
    56         if (v[1] > 0 && judge(root) && cal(root) != -1) cout << "Yes" << endl;
    57         else cout << "No" << endl;
    58     }
    59 }
     
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  • 原文地址:https://www.cnblogs.com/liuwenhan/p/11919723.html
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