PAT 1102 Invert a Binary Tree（树的遍历）

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

总结： 这道题目考察的就是树的遍历，层序遍历 + 中序遍历，经过前面几道遍历树的题目，这道题目还不算太难

代码：

#include <iostream>
#include <vector>
using namespace std;

struct node{
    int l,r;
}s[20];
vector<vector<int>> v(20);
vector<int> in;
void dfs(int root,int depth){
    v[depth].push_back(root);//将层序遍历的结果先保存在一个二维数组中
    
    if(s[root].r!=-1)   dfs(s[root].r,depth+1);
    in.push_back(root);//中序遍历结果保存在in数组中
    if(s[root].l!=-1)   dfs(s[root].l,depth+1);
}
int main(){
    int n;
    scanf("%d",&n);
    bool st[n]={false};
    
    char c;
    //输入数据
    for(int i=0;i<n;i++){
        getchar();
        scanf("%c",&c);
        if(c!='-'){
            s[i].l=c-'0';
            st[c-'0']=true;
        }
        else s[i].l=-1;
        getchar();
        scanf("%c",&c);
        if(c!='-'){
            s[i].r=c-'0';
            st[c-'0']=true;
        }
        else s[i].r=-1;
    }
    
    //找到根节点,遍历从根节点出发
    int root=0;
    while(st[root]) root++;
    dfs(root,0);
    for(int i=0;i<20;i++){
        for(int j=0;j<v[i].size();j++){
            if(i!=0)    printf(" ");
            printf("%d",v[i][j]);
        }
    }
    puts("");
    printf("%d",in[0]);
    for(int i=1;i<in.size();i++)    printf(" %d",in[i]);
    return 0;
}

柳神代码（可以学习一下）：

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

struct node{
    int id,l,r,index,level;
}s[110];
bool st[110];
vector<node> v1;
void dfs(int root,int index,int level){
    if(s[root].r!=-1)   dfs(s[root].r,index*2+2,level+1);
    v1.push_back({root,0,0,index,level});
    if(s[root].l!=-1)   dfs(s[root].l,index*2+1,level+1);
}
bool cmp(node a,node b){
    if(a.level!=b.level)    return a.level<b.level;//层与层之间不变
    else return a.index>b.index;//变化的是一个父节点的子节点  加入没有这个节点呢
}
int main(){
    int n;
    scanf("%d",&n);
    
    for(int i=0;i<n;i++){
        string l,r;
        s[i].id=i;
        cin >> l >> r;
        if(l!="-"){
            st[stoi(l)]=true;
            s[i].l=stoi(l);
        }
        else s[i].l=-1;
        if(r!="-"){
            st[stoi(r)]=true;
            s[i].r=stoi(r);
        }
        else s[i].r=-1;
    }
    int root=0;
    while(st[root])    root++;
    dfs(root,0,0);
    vector<node> v2(v1);//将数组v1赋值给v2
    sort(v2.begin(),v2.end(),cmp);
    for(int i=0;i<v2.size();i++){
        if(i!=0)    printf(" ");
        printf("%d",v2[i].id);
    }
    puts("");
    for(int i=0;i<v1.size();i++){
        if(i!=0)    printf(" ");
        printf("%d",v1[i].id);
    }
    return 0;
}

好好学习，天天向上！

我要考研！