1135 Is It A Red-Black Tree

题目：

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

思路：

1、红黑树也属于二叉搜索树（BST），只需要知道后序或前序遍历中的一个，就可以构造出树，具体的构造过程如下：

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;
}

for(int j = 0; j < n; j++){
    scanf("%d", &arr[j]);
    root = build(root, arr[j]);
}

2、性质5的判断：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

代码：

#include<stdio.h>
#include<iostream>
#include<math.h>
using namespace std;
struct node{
    int val; 
    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);
    }else{
        root -> right = build(root -> right, v);
    }
    return root;
}
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);
}
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);
}
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);
        node* root = NULL;      
        int arr[35];
        for(int j = 0; j < n; j++){
            scanf("%d", &arr[j]);
            root = build(root, arr[j]);
        }
        if(arr[0] < 0 || !judge1(root) || !judge2(root)){
            printf("No\n");
        }else{
            printf("Yes\n");
        }
    }
    return 0;
}