PAT 1099. Build A Binary Search Tree (30)

题目地址： http://www.patest.cn/contests/pat-a-practise/1099

单纯的数据结构题目

想到中序遍历的BST是有序的， 中序插入数， 层序遍历输出就可以了，这个题目为什么是30分呢， 纳闷

 1 #include<cstdio>

 2 #include<iostream>

 3 #include<queue>

 4 #include<algorithm>

 5 

 6 using namespace std;

 7 

 8 const int MAX_NODES = 1000;

 9 struct Node {

10         int lchild;

11         int rchild;

12         int num;

13 } nodes[MAX_NODES];

14 int len = 0;

15 

16 int nums[MAX_NODES];

17 int num_index = 0;

18 void in_order(int root) {

19         if (root != -1) {

20                 in_order(nodes[root].lchild);

21                 nodes[root].num = nums[num_index++];

22                 in_order(nodes[root].rchild);

23         }

24 }

25 

26 queue<int> q;

27 void level_order(int root) {

28         num_index = 0;

29         while(!q.empty()) q.pop(); //我不知道怎么清空队列， 只能这样了， 哈哈

30         q.push(root);

31         while(!q.empty()) {

32                 int x = q.front();

33                 q.pop();

34                 nums[num_index++] = nodes[x].num;

35                 if (nodes[x].lchild != -1)

36                         q.push(nodes[x].lchild);

37                 if (nodes[x].rchild != -1) 

38                         q.push(nodes[x].rchild);

39         }

40 }

41 

42 void debug_print() {

43         cout << "===============debug==========================" << endl;

44         for (int i = 0; i < len; ++i) {

45                 cout << nodes[i].num << endl;

46         }

47         cout << "================debug end =====================" << endl;

48 }

49 

50 int main() {

51         while(cin >> len) {

52                 for (int i = 0; i < len; ++i) {

53                         cin >> nodes[i].lchild >> nodes[i].rchild;

54                 }

55                 for (int i = 0; i < len; ++i) {

56                         cin >> nums[i];

57                 }

58                 sort(nums, nums + len);

59                 in_order(0);

60                 //debug_print();

61                 //cout << "post_order done" << endl;

62                 level_order(0);

63                 //cout << "level_order done" << endl;

64                 cout << nums[0];

65                 for (int i = 1; i < len; ++i) {

66                         cout << " " << nums[i];

67                 }

68                 cout << endl;

69         }

70         return 0;

71 }