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; }