1135 Is It A Red-Black Tree (30分)

最新推荐文章于 2021-06-20 23:44:05 发布

zbchenchanghao

最新推荐文章于 2021-06-20 23:44:05 发布

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本文探讨了红黑树的五种基本属性，并通过一个算法实现来验证给定的二叉搜索树是否符合红黑树的定义。红黑树是一种自平衡二叉查找树，具有节点颜色属性和平衡条件，确保树的高度大致对数增长。

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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

#include<iostream>

#include<vector>
#include<algorithm>
using namespace std;
struct node{

    struct node * lchild;
    struct node * rchild;
    int data;

};

node * build(node * root, int v) {
    if(root == NULL) {
        root = new node();
        root -> data = v;
        root -> lchild = root -> rchild = NULL;
    } else if(abs(v) <= abs(root->data))
        root -> lchild = build(root -> lchild, v);
    else
        root -> rchild = build(root -> rchild, v);
    return root;
}
node * creat_tree(vector<int> &pre_order, vector<int> &in_order, int pre_start, int pre_end, int in_start, int in_end){
// 用前序、中序建树会出现段错误，可能会有节点有相同的值
    if(pre_start > pre_end || in_start > in_end){

        return NULL;
    }

    node * root = new node;
    root -> data = pre_order[pre_start];
    int i;
    for(i = 0; i < in_order.size(); i++){

        if(abs(pre_order[pre_start]) == in_order[i])
            break;

    }
    root -> lchild = creat_tree(pre_order, in_order, pre_start + 1, pre_start + i - in_start, in_start, i - 1);
    root -> rchild = creat_tree(pre_order, in_order, pre_end - in_end + i + 1, pre_end, i + 1, in_end);


    return root;


}

bool is_child_pre(node * t){ //是否红节点的孩子节点是黑节点

    if(t == NULL)
        return true;
    if(t -> data < 0){
        if(t -> lchild != NULL && t -> lchild -> data < 0)
            return false;
        if(t -> rchild != NULL && t -> rchild -> data < 0)
            return false;
    }    

    return is_child_pre(t -> lchild) & is_child_pre(t -> rchild);

}

int black_level(node * t){

    if(t == NULL)
        return 0;
    int l = black_level(t -> lchild);
    int r = black_level(t -> rchild);
    return t -> data > 0 ? max(l, r) + 1 : max(l, r);
}

bool is_black_level(node * t){ //是否任意点到叶子节点黑节点个数（高度）相同

    if(t == NULL)
        return true;
    int l = black_level(t -> lchild);
    int r = black_level(t -> rchild);
    if(l != r)
        return false;
    return is_black_level(t -> lchild) & is_black_level(t -> rchild);
}

int main(){
    int k;
    cin >> k;
    for(int i = 0; i < k; i++){
        int n;
        scanf("%d", &n);
        vector<int> pre_order(n);
        // vector<int> in_order(n);
        node * tree = NULL;
        for(int j = 0; j < n; j++){

            scanf("%d", &pre_order[j]);
            // in_order[j] = abs(pre_order[j]);
            tree = build(tree, pre_order[j]);

        }
        // sort(in_order.begin(), in_order.end());
       
        
        // tree = creat_tree(pre_order, in_order, 0, n - 1, 0, n - 1);
        bool flag1 = true;
        bool flag2 = true;
        bool flag3 = true;
        if(pre_order[0] < 0){
            flag1 = false;
        }
        flag2 = is_child_pre(tree);
        flag3 = is_black_level(tree);
        if(flag1 && flag2 && flag3){
            printf("Yes\n");
        }else
        {
            printf("No\n");
        }
        







    }

    return 0;
}