1102 Invert a Binary Tree

题目来源：PAT (Advanced Level) Practice

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 (≤10) 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

words：

Invert 使反转，前后倒置 whiteboard 白板 adjacent 邻近的

题意：

将给定的一颗二叉树进行反转，即左右子树互换，然后输出层序遍历序列和中序遍历序列；

思路：

1. 本题的输入使以左子树右子树的形式，所以在输入时我们按照右子树左子树的形式输入，即可实现反转效果；

2. 用一个二维数组a[][2]来保存二叉树，行下标代表根结点，每行的两个元素分别代表右子树和左子树；

3. 用一个一维数组vis[]来保存一个结点是否为其他结点的孩子结点，下标代表结点；

4. 在vis[]中找出不是任何结点的孩子结点的结点，即根结点；

6. 从根结点开始进行层序遍历；

7. 从根结点开始进行中序遍历；

//PAT ad 1102 Invert a Binary Tree
#include <iostream>
using namespace std;
#define N 10
#include <queue>char a[N][2];
bool vis[N]={false};string ans_InOrder;  //前序遍历结果
string ans_level;       //层序遍历结果 void InOrder(int i)        //前序遍历
{if(a[i][0]!='-'&&a[i][1]!='-')   //左右子树均非空 {InOrder(a[i][1]-'0');      //递归遍历左子树 ans_InOrder+=to_string(i)+" ";   //输出当前根结点 InOrder(a[i][0]-'0');       //递归遍历右子树 }else if(a[i][1]!='-') //左子树非空 {InOrder(a[i][1]-'0');ans_InOrder+=to_string(i)+" ";}else if(a[i][0]!='-')    //右子树非空 {ans_InOrder+=to_string(i)+" ";InOrder(a[i][0]-'0'); }elseans_InOrder+=to_string(i)+" ";        //左右子都为空
}void Level(int i)  //层序遍历
{queue<int> que;que.push(i);  //压入根结点 while(!que.empty()){int x=que.front();que.pop();ans_level+=to_string(x)+" ";if(a[x][1]!='-')       //判断左子树 que.push(a[x][1]-'0');if(a[x][0]!='-')     //判断右子树 que.push(a[x][0]-'0');}
} int main()
{int n,i;cin>>n;char r,l;for(i=0;i<n;i++){cin>>r>>l;        //右子树、左子树 a[i][0]=r;a[i][1]=l;if(r>='0'&&r<='9')vis[r-'0']=true;   //标记是否为某个树的子节点 if(l>='0'&&l<='9')vis[l-'0']=true;}for(i=0;i<n;i++) //寻找根结点 if(vis[i]==false)break;Level(i);  //层序遍历 InOrder(i);  //前序遍历 ans_level.pop_back();    //舍弃末尾空格 ans_InOrder.pop_back();    cout<<ans_level<<endl;      //输出 cout<<ans_InOrder<<endl;   return 0;}