PAT (Advanced Level) Practice A1130 Infix Expression （25 分) (二叉樹、中序遍歷)

原題鏈接

#include <algorithm>
#include <iostream>
#include <fstream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <ctime>
#include <cstring>
#include <string>
#include <vector>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <deque>
using namespace std;

const int MAX = 30;

typedef struct BiTNode
{
    char data[12];
    bool is_root;
    struct BiTNode *lChild, *rChild;
}BiTNode, *BiTree;
BiTree T[MAX];
int N, L, R, root = 1;

vector<char*> V;
void inOrder(BiTree T, bool first)//先序遍歷
{
    if(!T) return ;
    if(!first && (T->lChild || T->rChild)) V.push_back("(");//最外層和葉節點不加括號，子樹加括號
    inOrder(T->lChild, 0);
    V.push_back(T->data);
    inOrder(T->rChild, 0);
    if(!first && (T->lChild || T->rChild)) V.push_back(")");
}

int main()
{
    scanf("%d", &N);
    for(int i=0; i<=N; i++)//初始化
    {
        T[i] = new BiTNode;
        T[i]->is_root = 1;
        T[i]->lChild = T[i]->rChild = NULL;
    }
    for(int i=1; i<=N; i++)//建樹
    {
        scanf("%s %d %d", &T[i]->data, &L, &R);
        if(L != -1)
        {
            T[i]->lChild = T[L];
            T[L]->is_root = 0;
        }
        if(R != -1)
        {
            T[i]->rChild = T[R];
            T[R]->is_root = 0;
        }
    }
    while(root <= N && !T[root]->is_root) root++;//查詢根節點
    inOrder(T[root], 1);//1標誌本次是最外層
    for(int i=0; i<V.size(); i++) printf("%s", V[i]);
    printf("\n");
	return 0;
}