【PAT】【Advanced Level】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

原题链接：

https://www.patest.cn/contests/pat-a-practise/1102

思路：

输入、建树。

遍历的时候，先右子树，再根节点、再左子树。同时记录层数、遍历次序

然后针对层序排序输出

最后输出中序序列

CODE：

#include<iostream>#include<cstring>#include<string>#include<cstdlib>#include<vector>#include<algorithm>#define N 11using namespace std;typedef struct S{int bh;int flo;int tra;int ls;int rs;};S t[N];bool fl[N];vector<S> ft;vector<S> it;void dfs(int n,int ff){if (t[n].rs!=-1){dfs(t[n].rs,ff+1);}t[n].tra=it.size();t[n].flo=ff;ft.push_back(t[n]);it.push_back(t[n]);if (t[n].ls!=-1){dfs(t[n].ls,ff+1);}}bool cmp(S a, S b){if (a.flo==b.flo){return a.tra<b.tra;}else{return a.flo<b.flo;}}int main(){memset(fl,0,sizeof(fl));int n;cin>>n;for (int i=0;i<n;i++){string a,b;cin>>a>>b;t[i].bh=i;if (a=="-")t[i].ls=-1;else{int temp=atoi(a.c_str());t[i].ls=temp;fl[temp]=1;}if (b=="-")t[i].rs=-1;else{int temp=atoi(b.c_str());t[i].rs=temp;fl[temp]=1;}}int root;for (int i=0;i<n;i++)if (fl[i]==0){root=i;break;}dfs(root,0);sort(ft.begin(),ft.end(),cmp);for (int i=0;i<n;i++){if (i!=0) cout<<" ";cout<<ft[i].bh;}cout<<endl;for (int i=0;i<n;i++){if (i!=0) cout<<" ";cout<<it[i].bh;}return 0;}