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 (

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 < 0) {
        //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 < 0) return false;
        if (root->right != NULL && root->right->val < 0) 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 0;//这里是返回个数0.
    int l = getNum(root->left);
    int r = getNum(root->right);
    return root->val > 0 ? max(l, r) + 1 : 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 = 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]);
        }
        //如果根节点不是红色，
        if (arr[0] < 0 || judge1(root) == false || judge2(root) == false)
            printf("No\n");
        else
            printf("Yes\n");
    }
    return 0;
}

//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。