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 1
6 5 7 4 3 2 0 1
这个题目已经将二叉树的结构信息给出了，题目的重点应该是在于invert，但是如果采用vector来保存节点的孩子信息，那么就不需要进行特意的Invert了，只要在遍历的时候，将vector中存储的两个孩子节点代表的意思取反即可，即将第一个作为右孩子，将第二个作为左孩子。然后就是使用广度优先搜索逐层打印信息，在使用中序遍历即可。
#include <iostream>#include <map>#include <vector>#include <queue>using namespace std;map<int,vector<int> >tree;vector<int> inor;void bfs(int root);void iner_traversal(int root);int main(){    int n;    cin >>n;    char tmp;    //输出二叉树的节点关系信息，并且找到根节点    int root_num;    vector<int> record(n,0);    for(int i=0;i<n;i++)    {        for(int j=0;j<2;j++)        {            cin >>tmp;            if(tmp=='-')                tree[i].push_back(-1);            else            {                tree[i].push_back(tmp-'0');                record[tmp-'0']++;            }        }    }    vector<int>::iterator iter=record.begin();    int i=0;    for(;iter!=record.end();iter++,i++)        if(*iter!=1)            root_num=i;    //广度优先搜索，逐层打印信息    bfs(root_num);    //中序遍历，将数值进行存储    iner_traversal(root_num);    //中序遍历结果输出    iter=inor.begin();    cout <<*iter;    iter++;    for(;iter!=inor.end();iter++)        cout <<" " <<*iter;    cout <<endl;    return 0;}void bfs(int root){    queue<int> q;    q.push(root);    bool flag=true;    while(!q.empty())    {        int tmp=q.front();        q.pop();        if(flag)        {            cout <<tmp;            flag=false;        }        else            cout <<" "<<tmp;        if(tree[tmp].at(1)!=-1)            q.push(tree[tmp].at(1));        if(tree[tmp].at(0)!=-1)            q.push(tree[tmp].at(0));    }    cout <<endl;}void iner_traversal(int root){    if(tree[root].at(1)!=-1)        iner_traversal(tree[root].at(1));    inor.push_back(root);    if(tree[root].at(0)!=-1)        iner_traversal(tree[root].at(0));}