A1102 Invert a Binary Tree

题目：https://pintia.cn/problem-sets/994805342720868352/problems/994805365537882112
bool数组赋值问题：https://blog.csdn.net/SmartLoveyu/article/details/100121485?depth_1-utm_source=distribute.pc_relevant.none-task&utm_source=distribute.pc_relevant.none-task

注意点

1. 这题主要是使用了静态链表
2. 代码写完找了半天bug，结果发现是isRoot数组的初始赋值问题，默认是赋false值的；下次直接赋值使用vector吧，或者赋false值
3. 另一个需要记住的是反转链表是后序遍历，先递归两个子再将它们反转。
4. 感觉今天对先序中序后序层序遍历已经有了一个基本的了解了

代码

#include<algorithm>
#include<iostream>
#include<queue>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn = 110;
struct Node{
    int lchild, rchild;
}node[maxn];
int n;
vector<bool> isRoot(maxn, true);
int strToNum(char c){
    if(c == '-') return -1;
    else {
        isRoot[c - '0'] = false;
        return c - '0';
    }
}
int find_root(){
    for(int i = 0; i < n; i++){
        if(isRoot[i] == true){
            return i;
        }
    }
}
void invert(int root){
    if(root == -1) return;
    invert(node[root].lchild);
    invert(node[root].rchild);
    swap(node[root].lchild, node[root].rchild);
}
void level(int root){
    int count = 0;
    queue<int> q;
    q.push(root);
    while(!q.empty()){
        int top = q.front();
        q.pop();
        if (count != 0) printf(" ");
        printf("%d",top);
        count++;
        if(node[top].lchild != -1) q.push(node[top].lchild);
        if(node[top].rchild != -1) q.push(node[top].rchild);
    }
}
int count_of_in = 0;
void inOrder(int root){
    if(root == -1) return;
    inOrder(node[root].lchild);
    if(count_of_in != 0) printf(" ");
    printf("%d",root);
    count_of_in++;
    inOrder(node[root].rchild);
}
int main(){
    char lchild, rchild;
    scanf("%d",&n);
    for(int i = 0; i < n; i++){
        //scanf("%c%c", &lchild, &rchild);
        cin>>lchild>>rchild;
        node[i].lchild = strToNum(lchild);
        node[i].rchild = strToNum(rchild);
    }
    int root = find_root();
    invert(root);
    level(root);
    cout<<endl;
    inOrder(root);
    return 0;
}