[PAT A1102]Invert a Binary Tree
题目描述
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!
输入格式
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.
输出格式
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.
输入样例
8
1 -
- -
0 -
2 7
- -
- -
5 -
4 6输出样例
3 7 2 6 4 0 5 1
6 5 7 4 3 2 0 1解析
1.题目大意：首先输入一个N，表示一棵树的结点的个数，并且默认编号为0~N-1，然后输入N行数据，第i行代表的是编号为i-1的结点，并且依次输出它的左儿子和它的右儿子编号。然后我们需要将其翻转，即对于所有结点而言，将其左右儿子交换，然后输出新树的层序遍历序列和中序遍历序列。
2.我是使用了数组来存储这样的一棵二叉树，因为编号是从0~N-1是连续的，故使用数组虽然跟一般的树结构不同(用到指针)，但是更好处理一些。
3.
#include<iostream>
#include<queue>
using namespace std;
const int maxn = 10; //结点个数不超过10
struct node {int parent = -1; //定义该结点的父节点编号int lchild = -1, rchild = -1; //该结点的左右儿子编号
};
void level_traverse(int root); //层序遍历函数
void in_traverse(int root); //中序遍历函数
node tree[maxn];
int n;
int main()
{int root = -1;cin >> n;getchar(); //去掉结尾的换行符for (int i = 0; i < n; i++) {char left, right;scanf("%c %c",&left, &right);getchar(); //去掉结尾的换行符if (left != '-') { //如果左儿子不为空tree[i].rchild = left - '0'; //这里定义“交换”操作，交换左右儿子tree[left - '0'].parent = i; //并且对于儿子结点的父亲结点编号是i}if (right != '-') {tree[i].lchild = right - '0'; //同理tree[right - '0'].parent = i;}}for (int i = 0; i < n; i++) {if (tree[i].parent == -1) { //这么多结点里面，没有父亲结点的结点就是根结点root = i;break;}}level_traverse(root);printf("\n");in_traverse(root);return 0;
}
void level_traverse(int root) {queue<int> que;que.push(root);while (!que.empty()) {int temp = que.front();printf("%d", temp);if (tree[temp].lchild != -1) que.push(tree[temp].lchild);if (tree[temp].rchild != -1) que.push(tree[temp].rchild);que.pop();if (!que.empty()) printf(" ");}
}
void in_traverse(int root) {if (root == -1) return;in_traverse(tree[root].lchild);printf("%d", root);n--;if (n != 0) printf(" ");in_traverse(tree[root].rchild);
}