红黑树简介与C++应用

简介

红黑树（Red Black Tree）是一种自平衡二叉查找树，是在计算机科学中用到的一种数据结构，典型的用途是实现关联数组。
它是在1972年由Rudolf Bayer发明的，当时被称为平衡二叉B树（symmetric binary B-trees）。后来，在1978年被 Leo J. Guibas 和 Robert Sedgewick 修改为如今的“红黑树”。
红黑树和AVL树类似，都是在进行插入和删除操作时通过特定操作保持二叉查找树的平衡，从而获得较高的查找性能。
它虽然是复杂的，但它的最坏情况运行时间也是非常良好的，并且在实践中是高效的：它可以在O(log n)时间内做查找，插入和删除，这里的n 是树中元素的数目。

红黑树的性质

性质1. 节点是红色或黑色。
性质2. 根节点是黑色。
性质3 每个叶节点（NIL节点，空节点）是黑色的。
性质4 每个红色节点的两个子节点都是黑色。(从每个叶子到根的所有路径上不能有两个连续的红色节点)
性质5. 从任一节点到其每个叶子的所有路径都包含相同数目的黑色节点。

PAT1135 Is It A Red-Black Tree

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.

在这里插入图片描述
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:

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<cmath>
using namespace std;
vector<int> arr;
struct node{
    int val;
    struct node *left, *right;//以链表的方式存储
};
 
node* build(node *root, int v){
    //递归二叉查找树
    if(root == NULL){
        root = new node();
        root->val = v;
        root->left = root->right = NULL;
    }else if(abs(v) <= abs(root->val))//小于根节点的为左边递归
        root->left = build(root->left, v);
    else
        root->right = build(root->right, v);
    return root;
}
 
bool judgeBlackChild(node *root){
    //判断红色节点的孩子为黑色
    if(root == NULL)return true;
    if(root->val < 0){//red negative
        if(root->left != NULL && root->left->val < 0 )return false;
        if(root->right != NULL && root->right->val < 0)return false;
    }
    return judgeBlackChild(root->left) && judgeBlackChild(root->right);
}
 
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);
    //大于0是黑色节点，如果是黑色节点则加1，否则不变
}
 
int judgeBlackNum(node *root){
    //判断数量是否相同
    if(root == NULL)return true;
    int l = getNum(root->left);
    int r = getNum(root->right);
    if(l != r)return false;
    return judgeBlackNum(root->left) && judgeBlackNum(root->right);
}
 
int main(){
    int k, n;
    cin>>k;
    for(int i = 0; i < k; i++){
        cin>>n;
        arr.resize(n);
        node *root = NULL;
        for(int j = 0; j < n; j++){
            cin>>arr[j];
            root = build(root, arr[j]);
        }
        if(arr[0] < 0 || !judgeBlackChild(root)|| !judgeBlackNum(root))
            cout<<"No\n";
        else
            cout<<"Yes\n";
    }
    return 0;
}

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