A1102. Invert a Binary Tree

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

 #include<cstdio>
 #include<iostream>
 #include<stack>
 #include<string.h>
 #include<queue>
 using namespace std;
 typedef struct NODE{
     int lchild, rchild;
     int data;
 }node;
 node tree[];
 int N, notRoot[] = {,};
 void postReverse(int root){
     if(root == -)
         return;
     if(tree[root].lchild != -)
         postReverse(tree[root].lchild);
     if(tree[root].rchild != -)
         postReverse(tree[root].rchild);
     swap(tree[root].lchild, tree[root].rchild);
 }
 void levelOrder(int root, int &cnt){
     queue<int> Q;
     if(root != -){
         Q.push(root);
     }
     while(Q.empty() == false){
         int index = Q.front();
         Q.pop();
         cnt++;
         if(cnt == N){
             printf("%d", tree[index].data);
         }else{
             printf("%d ", tree[index].data);
         }
         if(tree[index].lchild != -)
             Q.push(tree[index].lchild);
         if(tree[index].rchild != -)
             Q.push(tree[index].rchild);
     }
 }
 void inOrder(int root, int &cnt){
     if(root == -)
         return;
     if(tree[root].lchild != -)
         inOrder(tree[root].lchild, cnt);
     cnt++;
     if(cnt == N)
         printf("%d", tree[root].data);
     else printf("%d ", tree[root].data);
     if(tree[root].rchild != -)
         inOrder(tree[root].rchild, cnt);
 }
 int main(){
     char c1, c2;
     scanf("%d", &N);
     for(int i = ; i < N; i++){
         scanf("%*c%c %c", &c1, &c2);
         if(c1 == '-'){
             tree[i].lchild = -;
         }else{
             tree[i].lchild = c1 - '';
             notRoot[c1 - ''] = ;
         }
         if(c2 == '-'){
             tree[i].rchild = -;
         }else{
             tree[i].rchild = c2 - '';
             notRoot[c2 - ''] = ;
         }
         tree[i].data = i;
     }
     int root = ;
     for(int i = ; i < N; i++)
         if(notRoot[i] == ){
             root = i;
             break;
         }
     int cnt = ;
     postReverse(root);
     levelOrder(root, cnt);
     printf("\n");
     cnt = ;
     inOrder(root, cnt);
     cin >> N;
     return ;
 }

总结：

1、本题要求先对二叉树进行反转（左变成右），再层序和中序输出。由于二叉树的后序遍历是先访问左右子树，再访问根节点，与逆转具有相同的性质。逆转要求在左右子树都已经逆转之后，再将这两颗子树交换位置。因此逆转二叉树可以用后序遍历实现。

2、对于给数字编号、给出每个节点的左右孩子编号的输入数据，最好使用静态二叉树。将节点都保存在一个node数组中，仅仅对数组的下标进行操作。

3、静态二叉树寻找root：使用数组notRoot坐标记，在读入节点时，如果将其孩子节点标记为notRoot。输入完毕后遍历数组寻找root节点。

4、由于%c会将上一行的 \n 读入，所以每行之前要吸收 \n, 两个字符之间还要匹配空格。可以 scanf("%*c%c %c", &c1, &c2)； 其中%*c会读入一个字符，但被忽略，接收参数的是后两个。