PAT甲级——A1099 Build A Binary Search Tree

A Binary Search Tree (BST) is recursively defined as a binary tree which has the following properties:

The left subtree of a node contains only nodes with keys less than the node's key.
The right subtree of a node contains only nodes with keys greater than or equal to the node's key.
Both the left and right subtrees must also be binary search trees.

Given the structure of a binary tree and a sequence of distinct integer keys, there is only one way to fill these keys into the tree so that the resulting tree satisfies the definition of a BST. You are supposed to output the level order traversal sequence of that tree. The sample is illustrated by Figure 1 and 2.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer

Output Specification:

For each test case, print in one line the level order traversal sequence of that tree. All the numbers must be separated by a space, with no extra space at the end of the line.

Sample Input:

9
1 6
2 3
-1 -1
-1 4
5 -1
-1 -1
7 -1
-1 8
-1 -1
73 45 11 58 82 25 67 38 42

Sample Output:

58 25 82 11 38 67 45 73 42

 1 #include <iostream>
 2 #include <queue>
 3 #include <vector>
 4 #include <algorithm>
 5 using namespace std;
 6 struct Node
 7 {
 8     int val, l, r;
 9 }node[105];
10 vector<int>nums(105), levelOrder;
11 int N, k = 0;
12 void inOrderTravel(int root)//得到树的中序遍历
13 {
14     if (root == -1)
15         return;
16     inOrderTravel(node[root].l);
17     node[root].val = nums[k++];
18     inOrderTravel(node[root].r);
19 }
20 void levelOrderTravel(int root)//得到树的中序遍历
21 {
22     queue<int>q;
23     q.push(root);
24     while (!q.empty())
25     {
26         root = q.front();
27         q.pop();
28         levelOrder.push_back(node[root].val);
29         if (node[root].l != -1)
30             q.push(node[root].l);
31         if (node[root].r != -1)
32             q.push(node[root].r);
33     }
34 }
35 int main()
36 {
37     cin >> N;    
38     int l, r;
39     int root = 0;
40     for (int i = 0; i < N; ++i)//按题目意思使用前序遍历构建一棵树
41     {
42         cin >> l >> r;
43         node[i].l = l;
44         node[i].r = r;
45     }
46     for (int i = 0; i < N; ++i)
47         cin >> nums[i];
48     sort(nums.begin(), nums.begin() + N);//得到中序遍历
49     inOrderTravel(root);//通过中序遍历重构二叉树
50     levelOrderTravel(root);
51     for (int i = 0; i < N; ++i)
52         cout << levelOrder[i] << (i == N - 1 ? "" : " ");
53     return 0;
54 }