1102 Invert a Binary Tree（树的遍历 dfs/bfs）

1102 Invert a Binary Tree （25 分）

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

解题思路：

本题涉及树的广度优先遍历(bfs，层序遍历)和深度优先遍历（中序遍历），此处题面提及反转二叉树，这里只需在输入环节将输入的数据反转（第一个数给右子树，第二个数给左子树）。

#include <iostream>
#include <vector>
#include <queue>
using namespace std;
struct node
{
    int r,l;
};
vector<node>v;
void bfs(int root){
    queue<int>q;
    q.push(root);
    bool first1=true;
    while(!q.empty()){
        int u=q.front();
        if(first1){
            cout<<u;
            first1=false;
        }else cout<<" "<<u;
        q.pop();
        if(v[u].l!=-1) q.push(v[u].l);
        if(v[u].r!=-1) q.push(v[u].r);
    }
    cout<<endl;
}
bool isfirst2=true;
void in_travel(int u){
    if(v[u].l!=-1) in_travel(v[u].l);
    if(isfirst2) {
        cout<<u;
        isfirst2=false;
    }else{
        cout<<" "<<u;
    }
    if(v[u].r!=-1) in_travel(v[u].r);
}
int main(){
    int n;
    cin>>n;
    v.resize(n);
    vector<bool> exist(n,false);
    for(int i=0;i<n;i++){
        char x,y;
        cin>>x>>y;
        if(x=='-')v[i].r=-1;
        else {
            v[i].r=x-'0';
            exist[v[i].r]=true;
        }
        if(y=='-') v[i].l=-1;
        else{
            v[i].l=y-'0';
            exist[v[i].l]=true;
        }
    }
    int root=0;
    while(root<n && exist[root]==true) root++;
    bfs(root);
    in_travel(root);
    cout<<endl;
    return 0;
}