1102 Invert a Binary Tree (25 point(s))

1102 Invert a Binary Tree (25 point(s))

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

题意：给定一棵二叉树，给出其反转树的层次遍历、中序遍历。

考察点：树的静态存储、寻找树的根、树的反转、树的遍历

思路：

1. 静态建立二叉树，同时标记儿子节点。完成后根据标记结果（有且仅有根节点没有被标记）找出根；

2. 由于要的是反转二叉树，存储时左子树、右子树反着存储即可；

3. 层次遍历（BFS）；

4. 中序遍历（递归）。

#include<iostream>
#include<queue>
#include<vector>
#include<algorithm>
#include<cstring>
using namespace std;
struct Node{
	int left=-1;int right=-1;
};
Node node[17];
bool visit[17]={false};
int string2int(string s){
	int ans = 0;
	for(int i=0;i<s.length();i++){
		ans = ans*10+s[i]-'0';
	}
	return ans;
}
void levelOrder(int root,vector<int> &v){
	queue<int> q;
	q.push(root);
	while(!q.empty()){
		int top = q.front();
		q.pop();
		v.push_back(top);
		if(node[top].left!=-1) q.push(node[top].left);
		if(node[top].right!=-1) q.push(node[top].right);
	}
} 
void output(vector<int> &v){
	for(int i=0;i<v.size();i++){
		if(i==0) cout<<v[i];
		else cout<<" "<<v[i];
	}
	cout<<endl;
	v.clear();
}
void inOrder(int root,vector<int> &s){
	if(node[root].left!=-1) inOrder(node[root].left,s);
	s.push_back(root);
	if(node[root].right!=-1) inOrder(node[root].right,s);
} 
int main(void){
	int N;cin>>N;
	string a,b;
	for(int i=0;i<N;i++){
		cin>>a>>b;
		if(a!="-"){//反着存储 
			node[i].right = string2int(a);
			visit[node[i].right] = true;
		}
		if(b!="-"){
			node[i].left = string2int(b);
			visit[node[i].left] = true;
		}
	}
	int root = find(visit,visit+N,false)-visit;
	vector<int> s;
	levelOrder(root,s);
	output(s);
	inOrder(root,s);
	output(s);
	return 0;
}