PAT_A1102#Invert a Binary Tree

Source:

PAT A1102 Invert a Binary Tree (25 分)

Description:

The following is from Max Howell @twitter:

Google: 90% of our engineers use the software you wrote (Homebrew), but you can't invert a binary tree on a whiteboard so fuck off.

Now it's your turn to prove that YOU CAN invert a binary tree!

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (≤) which is the total number of nodes in the tree -- and hence the nodes are numbered from 0 to N−1. Then N lines follow, each corresponds to a node from 0 to N−1, and gives the indices of the left and right children of the node. If the child does not exist, a - will be put at the position. Any pair of children are separated by a space.

Output Specification:

For each test case, print in the first line the level-order, and then in the second line the in-order traversal sequences of the inverted tree. There must be exactly one space between any adjacent numbers, and no extra space at the end of the line.

Sample Input:

8
1 -
- -
0 -
2 7
- -
- -
5 -
4 6

Sample Output:

3 7 2 6 4 0 5 1
6 5 7 4 3 2 0 1

Keys:

Code:

 1 /*
 2 time: 2019-06-30 14:09:56
 3 problem: PAT_A1102#Invert a Binary Tree
 4 AC: 23:00
 5 
 6 题目大意：
 7 打印镜像树层序和中序遍历
 8 输入：
 9 第一行给出，结点数N<=10
10 接下来N行，结点i（0~n-1）的左孩子和右孩子
11 
12 基本思路：
13 构造静态树遍历
14 */
15 #include<cstdio>
16 #include<queue>
17 #include<string>
18 #include<iostream>
19 using namespace std;
20 const int M=1e2;
21 int mp[M]={0},n;
22 struct node
23 {
24     int lchild,rchild;
25 }tree[M];
26 
27 void LayerOrder(int root)
28 {
29     queue<int> q;
30     q.push(root);
31     int pt=0;
32     while(!q.empty())
33     {
34         root = q.front();
35         q.pop();
36         printf("%d%c", root, ++pt==n?'\n':' ');
37         if(tree[root].rchild != -1)
38             q.push(tree[root].rchild);
39         if(tree[root].lchild != -1)
40             q.push(tree[root].lchild);
41     }
42 }
43 
44 void InOrder(int root)
45 {
46     if(root == -1)
47         return;
48     static int pt=0;
49     InOrder(tree[root].rchild);
50     printf("%d%c", root, ++pt==n?'\n':' ');
51     InOrder(tree[root].lchild);
52 }
53 
54 int main()
55 {
56 #ifdef ONLINE_JUDGE
57 #else
58     freopen("Test.txt", "r", stdin);
59 #endif // ONLINE_JUDGE
60 
61     scanf("%d", &n);
62     string r,l;
63     for(int i=0; i<n; i++){
64         cin >> l >> r;
65         if(l == "-")
66             tree[i].lchild = -1;
67         else{
68             tree[i].lchild = atoi(l.c_str());
69             mp[tree[i].lchild]=1;
70         }
71         if(r == "-")
72             tree[i].rchild = -1;
73         else{
74             tree[i].rchild = atoi(r.c_str());
75             mp[tree[i].rchild]=1;
76         }
77     }
78     int root;
79     for(int i=0; i<n; i++)
80         if(mp[i]==0)
81             root=i;
82     LayerOrder(root);
83     InOrder(root);
84 
85     return 0;
86 }

 

posted @ 2019-06-30 14:12  林東雨  阅读(150)  评论(0)  编辑  收藏  举报