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

对于这种输入，多半是建静态树，画下树的图，很容易发现规律。

#include<bits/stdc++.h>
using namespace std;
struct node{
    int data,lchild,rchild;
}T[100];
int k=0;
void Inorder(int root){
    if(root==11)return ;
    Inorder(T[root].lchild);
    if(k++)cout<<' ';
    cout<<T[root].data;
    Inorder(T[root].rchild);
}
void levelorder(int root){
    queue<int>q;
    q.push(root);
    int i=0;
    while(!q.empty()){
        int root=q.front();
        q.pop();
        if(i++)cout<<' ';
        cout<<root;
        if(T[root].lchild!=11)  q.push(T[root].lchild);
        if(T[root].rchild!=11)  q.push(T[root].rchild);
    }
}
int G[15][15];
int main(){
    int N;
    cin>>N;
    for(int i=0;i<N;i++){
        char l,r;
        getchar();
        scanf("%c %c",&r,&l);
        if(l!='-'){
            T[i].lchild=l-'0';
            G[i][l-'0']=1;
        }
        else T[i].lchild=11;
        if(r!='-'){
            T[i].rchild=r-'0';
            G[i][r-'0']=1;  
        }
        else T[i].rchild=11;
        T[i].data=i;    
    }
    int root;
    for(int i=0;i<N;i++){
        int is=1;
        for(int j=0;j<N;j++)
            if(G[j][i]!=0)is=0;
        if(is)root=i;
    }
    levelorder(root);
    cout<<endl;
    Inorder(root);
    return 0;
}