PAT 1135 Is It A Red-Black Tree[难]

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

题目大意：给出了红黑树的定义，并且给出n组样例，并且每一组给除了先根遍历结果，判断每一组是否是红黑树，红色节点用负号表示。

//可以用先根遍历来确定一棵二叉树吗？

1.根节点是黑色

2.如果一个节点是黑色，那么子节点为红色

3.在所有从根节点到叶节点的路径上，黑色节点的个数相同。

红黑树也是一个二叉搜索树，所以能够根据前序来建树！！　　

代码转自：https://www.liuchuo.net/archives/4099

#include <iostream>
#include <vector>
#include <cmath>
#include<cstdio>
using namespace std;
vector<int> arr;
struct node {
    int val;
    struct node *left, *right;
};
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);//递归建树！！！
    else
        root->right = build(root->right, v);
    return root;
}
bool judge1(node *root) {
    if (root == NULL) return true;
    if (root->val < ) {
        //如果当前是红色节点，去判断左右子树是否是黑色节点。
        if (root->left != NULL && root->left->val < ) return false;
        if (root->right != NULL && root->right->val < ) return false;
    }
    //再递归地去判断左子节点的子数，和右子节点的子数。
    return judge1(root->left) && judge1(root->right);
}
int getNum(node *root) {
    if (root == NULL) return ;//这里是返回个数0.
    int l = getNum(root->left);
    int r = getNum(root->right);
    return root->val >  ? max(l, r) +  : max(l, r);
    //如果根节点是黑节点，那么+1.
}
bool judge2(node *root) {
    if (root == NULL) return true;
    int l = getNum(root->left);//获取左右子树的黑节点个数。
    int r = getNum(root->right);
    if(l != r) return false;//图三在根节点调用时在此处就会返回false。
    return judge2(root->left) && judge2(root->right);//再去递归判断子树中是否符合，
    //像图二就会在这次递归中判断为false.
}
int main() {
    int k, n;
    scanf("%d", &k);
    for (int i = ; i < k; i++) {
        scanf("%d", &n);
        arr.resize(n);
        node *root = NULL;
        for (int j = ; j < n; j++) {
            scanf("%d", &arr[j]);
            root = build(root, arr[j]);
        }
        //如果根节点不是红色，
        if (arr[] <  || judge1(root) == false || judge2(root) == false)
            printf("No\n");
        else
            printf("Yes\n");
    }
    return ;
}

//柳神真厉害。

1.二叉树这类问题都是使用递归去做的，掌握递归的思想十分重要。

2.根据条件去写出函数来判断。

3.要多复习。