1102 Invert a Binary Tree （25 分）dfs+层序+中序+后序遍历

1102 Invert a Binary Tree （25 分）

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

思路：　　这个题目的难点是寻找树根，我的办法是使用dfs，从哪一个结点出发能够访问到所有的结点，那么该结点就是树根。可以认为存储的是一个有n-1条边的有向图，只有从根节点出发才能到达所有的结点。然后使用后序遍历进行镜像反转，按层序和后序输出即可。

#include<iostream>
#include<vector>
#include<algorithm>
#include<queue>
#include<string>
#include<map>
#include<set>
using namespace std;

vector<int> visited;

struct Node
{

    int lchild,rchild;
};
Node node[11];
void findRoot(int root)
{
    if(root==-1)
        return;
    visited.push_back(root);
    findRoot(node[root].lchild);
    findRoot(node[root].rchild);
}

void invert(int root)
{
    if(root!=-1)
    {
        invert(node[root].lchild);
        invert(node[root].rchild);
        swap(node[root].lchild,node[root].rchild);
    }
}

void levelOrder(int root)
{
    queue<int>qu;
    qu.push(root);
    cout<<root;
    while(!qu.empty())
    {
        int temp=qu.front();
        qu.pop();
        if(node[temp].lchild!=-1)
        {
            qu.push(node[temp].lchild);
            cout<<" "<<node[temp].lchild;
        }
        if(node[temp].rchild!=-1)
        {
            qu.push(node[temp].rchild);
            cout<<" "<<node[temp].rchild;
        }
    }
}
vector<int> in_order;
void inOrder(int root)
{
    if(root!=-1)
    {
        inOrder(node[root].lchild);
        in_order.push_back(root);
        inOrder(node[root].rchild);
    }
}

int main()
{
    int n;
    cin>>n;

    for(int i=0;i<n;i++)
    {
        string temp1,temp2;
        cin>>temp1>>temp2;
        if(temp1[0]==‘-‘)
            node[i].lchild=-1;
        else
            node[i].lchild=atoi(temp1.c_str());
        if(temp2[0]==‘-‘)
            node[i].rchild=-1;
        else
            node[i].rchild=atoi(temp2.c_str());
    }
    int i;
    for(i=0;i<n;i++)
    {
        findRoot(i);
        if(visited.size()==n)
            break;
        else
            visited.clear();
    }
    //cout<<i<<endl;
    invert(i);
    levelOrder(i);
    inOrder(i);
    cout<<endl;
    cout<<in_order[0];
    for(int i=1;i<in_order.size();i++)
        cout<<" "<<in_order[i];
    return 0;
}

原文地址：https://www.cnblogs.com/zhanghaijie/p/10306358.html