【笨方法学PAT】1102 Invert a Binary Tree （25 分）

一、题目

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 Nlines 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

二、题目大意

给出原二叉树的每个结点的左右孩子，反转一棵二叉树，输出它的层序和前序遍历。

三、考点

树、DFS

四、注意

1、使用struct node{}建树，注意 - 的处理，可以将树左右孩子初始化-1；

2、递归反转；

3、递归获得中序。

五、代码

#include<iostream>
#include<vector>
#include<string>
#include<queue>
#include<algorithm>
using namespace std;
struct node {
	int val;
	int left=-1, right=-1;
};
vector<node> vec;
vector<int> vec_in;

void invertTree(int root) {
	if(vec[root].left!=-1)
		invertTree(vec[root].left);
	if (vec[root].right != -1)
		invertTree(vec[root].right);
	swap(vec[root].left, vec[root].right);
}

void dfs(int root) {
	if (vec[root].left != -1)
		dfs(vec[root].left);
	vec_in.push_back(root);
	if (vec[root].right != -1)
		dfs(vec[root].right);
}
int main() {
	//read
	int n;
	cin >> n;
	vec.resize(n);
	vector<bool> show(n, false);

	//read and build tree
	for (int i = 0; i < n; ++i) {
		string s1, s2;
		cin >> s1 >> s2;
		if (s1 != "-") {
			vec[i].left = stoi(s1);
			show[stoi(s1)] = true;
		}
		if (s2 != "-") {
			vec[i].right = stoi(s2);
			show[stoi(s2)] = true;
		}
	}

	//find root
	int root;
	for (int i = 0; i < n; ++i) {
		if (show[i] == false) {
			root = i;
			break;
		}
	}

	//invert
	invertTree(root);

	//level order
	vector<int> vec_level;
	queue<int> que;
	que.push(root);
	while (!que.empty()) {
		int r = que.front();
		que.pop();
		vec_level.push_back(r);
		if (vec[r].left != -1)
			que.push(vec[r].left);
		if (vec[r].right != -1)
			que.push(vec[r].right);
	}

	//output level
	for (int i = 0; i < n; ++i) {
		if (i != 0)
			cout << " ";
		cout << vec_level[i];
	}
	cout << endl;

	//in-order
	dfs(root);

	//output in
	for (int i = 0; i < n; ++i) {
		if (i != 0)
			cout << " ";
		cout << vec_in[i];
	}

	system("pause");
	return 0;
}