17年春季第四题 PAT甲级 1135 Is It A Red-Black Tree (30)

水深无声 2023-07-13 04:26 2阅读 0赞

题目来源：[https://pintia.cn/problem-sets/994805342720868352/problems/994805346063728640][https_pintia.cn_problem-sets_994805342720868352_problems_994805346063728640]

备考汇总贴：[2020年3月PAT甲级满分必备刷题技巧][2020_3_PAT]

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.

<table> 
 <thead> 
  <tr> 
   <th><img alt="rbf1.jpg" src="https://imgconvert.csdnimg.cn/aHR0cHM6Ly9pbWFnZXMucHRhdXNlcmNvbnRlbnQuY29tL2VmZjgwYmQ0LWM4MzMtNDgxOC05Nzg2LTgxNjgwZDFiMzA0YS5qcGc?x-oss-process=image/format,png"></th> 
   <th><img alt="rbf2.jpg" src="https://imgconvert.csdnimg.cn/aHR0cHM6Ly9pbWFnZXMucHRhdXNlcmNvbnRlbnQuY29tL2IxMTE4NGRmLWVhYWItNDUxYy1iN2Q0LTdmYzFkYzgyYjAyOC5qcGc?x-oss-process=image/format,png"></th> 
   <th><img alt="rbf3.jpg" src="https://imgconvert.csdnimg.cn/aHR0cHM6Ly9pbWFnZXMucHRhdXNlcmNvbnRlbnQuY29tLzYyNWM1MzJiLTIyZmMtNDdiOS04MGVhLTA1MzdjZjAwZDkyMi5qcGc?x-oss-process=image/format,png"></th> 
  </tr> 
 </thead> 
 <tbody> 
  <tr> 
   <td>Figure 1</td> 
   <td>Figure 2</td> 
   <td>Figure 3</td> 
  </tr> 
 </tbody> 
</table>

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（以下内容主要转载自[liuchuo][]，经过了我的编辑补充） #

给一棵二叉搜索树的前序遍历，判断它是否为红黑树，是输出Yes，否则输出No。

# 题目分析 #

一、这道题必须按照题目给的红黑树五个定义去思考：  
(1) Every node is either red or black. （不用判断）  
(2) The root is black.（需要判断）  
(3) Every leaf (NULL) is black.（不用判断，但是可以用在4的判断里）  
(4) If a node is red, then both its children are black.（需要判断）  
(5) For each node, all simple paths from the node to descendant leaves contain the same number of black nodes.（需要判断）

题目还提到红黑树是一种二叉搜索树（所以中序唯一确定），那给定前序就能确定一棵树了。

题目还说红黑树是“balanced”（平衡的），不用管。

二、整理一下，整个程序要做的是判断以下几点：  
1.根结点是否为黑色  
2.如果一个结点是红色，它的孩子节点是否都为黑色  
3.从任意结点到叶子结点的路径中，黑色结点的个数是否相同  
所以分为以下几步：  
0. 根据先序建立一棵树，用链表表示（常规的BST的建树方法）  
1. 判断根结点（题目所给先序的第一个点即根结点）是否是黑色  
2. 根据建立的树，从根结点开始遍历，如果当前结点是红色，判断它的孩子节点是否为黑色，递归返回结果  
3. 从根节点开始，递归遍历，检查每个结点的左子树的高度和右子树的高度（这里的高度指黑色结点的个数），比较左右孩子高度是否相等，递归返回结果

# 满分代码 #

#include <iostream>
    #include <vector>
    #include <cmath>
    using namespace std;
    vector<int> arr;
    struct node {
        int val;
        struct node *left, *right;
    };
    //建树
    node* build(node *root, int v) {
        if(root == NULL) {
            root = new node();
            root->val = v;
            root->left = root->right = NULL;
        } else if(abs(v) <= abs(root->val))
            root->left = build(root->left, v);
        else
            root->right = build(root->right, v);
        return root;
    }
    //从根结点开始遍历，如果当前结点是红色，判断它的孩子节点是否为黑色，递归返回结果
    bool judge1(node *root) {
        if (root == NULL) return true;
        if (root->val < 0) {
            if (root->left != NULL && root->left->val < 0) return false;
            if (root->right != NULL && root->right->val < 0) return false;
        }
        return judge1(root->left) && judge1(root->right);
    }
    //求出节点的高度（这里的高度指黑色结点的个数）
    int getNum(node *root) {
        if (root == NULL) return 0;
        int l = getNum(root->left);
        int r = getNum(root->right);
        return root->val > 0 ? max(l, r) + 1 : max(l, r);
    }
    //检查每个结点的左子树的高度和右子树的高度（这里的高度指黑色结点的个数），比较左右孩子高度是否相等，递归返回结果
    bool judge2(node *root) {
        if (root == NULL) return true;
        int l = getNum(root->left);
        int r = getNum(root->right);
        if(l != r) return false;
        return judge2(root->left) && judge2(root->right);
    }
    int main() {
        int k, n;
        scanf("%d", &k);
        for (int i = 0; i < k; i++) {
            scanf("%d", &n);
            arr.resize(n);
            node *root = NULL;
            for (int j = 0; j < n; j++) {
                scanf("%d", &arr[j]);
                root = build(root, arr[j]);
            }
            if (arr[0] < 0 || judge1(root) == false || judge2(root) == false)
                printf("No\n");
            else
                printf("Yes\n");
        }
        return 0;
    }

[https_pintia.cn_problem-sets_994805342720868352_problems_994805346063728640]: https://pintia.cn/problem-sets/994805342720868352/problems/994805346063728640
[2020_3_PAT]: https://blog.csdn.net/allisonshing/article/details/104213959
[liuchuo]: https://blog.csdn.net/liuchuo/article/details/78037334