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:

• 二叉树的建立和遍历

Code:

 /*
 time: 2019-06-30 14:09:56
 problem: PAT_A1102#Invert a Binary Tree
 AC: 23:00

 题目大意：
 打印镜像树层序和中序遍历
 输入：
 第一行给出，结点数N<=10
 接下来N行，结点i（0~n-1）的左孩子和右孩子

 基本思路：
 构造静态树遍历
 */
 #include<cstdio>
 #include<queue>
 #include<string>
 #include<iostream>
 using namespace std;
 const int M=1e2;
 int mp[M]={},n;
 struct node
 {
     int lchild,rchild;
 }tree[M];

 void LayerOrder(int root)
 {
     queue<int> q;
     q.push(root);
     int pt=;
     while(!q.empty())
     {
         root = q.front();
         q.pop();
         printf("%d%c", root, ++pt==n?'\n':' ');
         if(tree[root].rchild != -)
             q.push(tree[root].rchild);
         if(tree[root].lchild != -)
             q.push(tree[root].lchild);
     }
 }

 void InOrder(int root)
 {
     if(root == -)
         return;
     static int pt=;
     InOrder(tree[root].rchild);
     printf("%d%c", root, ++pt==n?'\n':' ');
     InOrder(tree[root].lchild);
 }

 int main()
 {
 #ifdef ONLINE_JUDGE
 #else
     freopen("Test.txt", "r", stdin);
 #endif // ONLINE_JUDGE

     scanf("%d", &n);
     string r,l;
     for(int i=; i<n; i++){
         cin >> l >> r;
         if(l == "-")
             tree[i].lchild = -;
         else{
             tree[i].lchild = atoi(l.c_str());
             mp[tree[i].lchild]=;
         }
         if(r == "-")
             tree[i].rchild = -;
         else{
             tree[i].rchild = atoi(r.c_str());
             mp[tree[i].rchild]=;
         }
     }
     int root;
     for(int i=; i<n; i++)
         if(mp[i]==)
             root=i;
     LayerOrder(root);
     InOrder(root);

     return ;
 }