1102. Invert a Binary Tree (25)-蒲公英云

1102. Invert a Binary Tree (25)

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

题目大意：

代码：

#include<stdio.h>
#include<queue>
using namespace std;
struct node
{
    int flag;
    int lchild,rchild;
}tree[20];
int flag=0;
void inorder(int root)
{
    //printf("%d\n",root);
    if(root!=-1)
    {
        inorder(tree[root].rchild);
        if(flag==0)
        {
            printf("%d",root);
            flag=1;
        }
        else
        {
            printf(" %d",root);
        }
        inorder(tree[root].lchild);
    }
}
int tag=0;
void levelorder(int root)
{
    queue<int> q;
    q.push(root);
    while(!q.empty())
    {
        int tmp=q.front();
        q.pop();
        if(tmp!=-1)
        {
            q.push(tree[tmp].rchild);
            q.push(tree[tmp].lchild);
            if(tag==0)
            {
               printf("%d",tmp);
               tag=1;
            }
            else
            {
               printf(" %d",tmp);
            }
        }
    }
}
int main()
{
    int i,j,n,m,k,t;
    char l[2],r[2];
    scanf("%d",&n);
    for(i=0;i<n;i++)
    {
        scanf("%s %s",l,r);
        if(l[0]!='-')
        {
            tree[i].lchild=l[0]-'0';
            tree[l[0]-'0'].flag=1;
        }
        else
        {
            tree[i].lchild=-1;
        }
        if(r[0]!='-')
        {
            tree[i].rchild=r[0]-'0';
            tree[r[0]-'0'].flag=1;
        }
        else
        {
            tree[i].rchild=-1;
        }
    }
    for(i=0;i<n;i++)
    {
        if(tree[i].flag==0)
        {
            k=i;
            break;
        }
    }
    //printf("%d\n",k);
    levelorder(k);
    printf("\n");
    inorder(k);
    printf("\n");
return 0;
}