PAT 甲级 1102 Invert a Binary Tree (25分)

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本文详细介绍了二叉树的静态写法及遍历算法，包括先序、中序和后序遍历，重点讲解了如何通过层级遍历和中序遍历来获取二叉树的节点序列。同时，提供了完整的C++代码实现，帮助读者深入理解二叉树的遍历过程。

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

二叉树的静态写法: 
        struct node{
            int val;
            int left;
            int right;
        } 

一般树的静态写法:
        struct node{
            int val;
            vector<int> child;
        }

树中的结点用一个node[]数组来盛放

#include<iostream>
#include<queue>
#include<vector>
using namespace std;
struct node{
    int left = -1;
    int right = -1;
};
node nodes[20];
int st[20];
queue<int> q;
vector<int> ans;
void inorder(int h){
    if (h == -1) return;
    inorder(nodes[h].right);
    ans.push_back(h);
    inorder(nodes[h].left);
}

void level(int h){
    q.push(h);
    while(!q.empty()){
        int t = q.front();
        q.pop();
        ans.push_back(t);
        if (nodes[t].right != -1) q.push(nodes[t].right);
        if (nodes[t].left != -1)  q.push(nodes[t].left);
    }
}

int main(){

    int n,head; char id1,id2;
    cin>>n;
    for (int i = 0; i < n; ++i){
        cin>>id1>>id2;
        if (id1 != '-') {
            nodes[i].left = (id1 - '0');
            st[id1 - '0'] ++;
        }
        if (id2 != '-'){
            nodes[i].right = (id2 - '0');
            st[id2 - '0'] ++;
        }
    }
    for (int i = 0; i < n; ++i)  // 寻找头结点
        if (st[i] == 0){
            head = i;
            break;
        }
    level(head);
    for (int i = 0; i < ans.size(); ++i){
        if (i) cout<<" "<<ans[i];
        else cout<<ans[i];
    }
    ans.clear();
    cout<<endl;
    inorder(head);
    for (int i = 0; i < ans.size(); ++i){
        if (i) cout<<" "<<ans[i];
        else cout<<ans[i];
    }
    return 0;
}