PAT 1102. Invert a Binary Tree (25)

1102. Invert a Binary Tree (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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:

81 -- -0 -2 7- -- -5 -4 6

Sample Output:

3 7 2 6 4 0 5 16 5 7 4 3 2 0 1

坑点在于给的数据是先右孩子再左孩子。

#include <iostream>#include<stdio.h>#include<string>#include<queue>#include<algorithm>#include<string.h>#include<vector>#include<map>#include<math.h>using namespace std;struct{    int lc,rc;    int now;}node[15];int mark[15]={0};void inorder(int xx) {     static int flag=0;     if(node[xx].lc!=-1) inorder(node[xx].lc);     if(flag==1) cout<<' ';     flag=1;     cout<<xx;     if(node[xx].rc!=-1) inorder(node[xx].rc); }void levelorder(int x) {     queue<int> Q;     Q.push(x);     int flag=0;     while(!Q.empty())     {         int tmp=Q.front();         Q.pop();        if(flag==1) cout<<' ';        flag=1;        cout<<tmp;        if(node[tmp].lc!=-1) Q.push(node[tmp].lc);        if(node[tmp].rc!=-1) Q.push(node[tmp].rc);     }     cout<<endl; } int main(){    int n;    cin>>n;    char a[5],b[5];    for(int i=0;i<n;i++)    {        cin>>a>>b;        if(a[0]=='-') node[i].lc=-1;        else        {        node[i].lc=atoi(a);        mark[atoi(a)]++;        }        if(b[0]=='-') node[i].rc=-1;        else        {            mark[atoi(b)]++;            node[i].rc=atoi(b);        }        swap(node[i].lc,node[i].rc);        node[i].now=i;    }     int start;     for(int i=0;i<n;i++)        if(mark[i]==0)     {         start=i;break;     }     levelorder(start);     inorder(start);  return 0;}