哇，这题写的都是泪啊，一开始想估计数据比较水就直接用树状数组模拟了，然后#4一直re，但是只跑了30ms，4千万点数组现在30ms就跑完了？？？然后放弃了直接用vector的。。。

#include <iostream>
#include <vector>
using namespace std;
typedef struct node{
    char value;
    int leftChild;
    int rightChild;
}Node;
vector<Node>tree;
void A(int num){
    if(num!=-1&&tree[num].value!=-1){
        cout<<tree[num].value;
        A(tree[num].leftChild);
        A(tree[num].rightChild);
    }
}
void B(int num){
    if(num!=-1&&tree[num].value!=-1){
        B(tree[num].leftChild);
        cout<<tree[num].value;
        B(tree[num].rightChild);
    }
}
void C(int num){
    if(num!=-1&&tree[num].value!=-1){
        C(tree[num].leftChild);
        C(tree[num].rightChild);
        cout<<tree[num].value;
    }
}
int main(void){
    char head,temp;
    int pos=0,end=0;
    bool flag=0;
    cin>>head;
    getchar();
    Node* ptr=new Node;
    ptr->value=head;
    ptr->leftChild=-1;
    ptr->rightChild=-1;
    tree.push_back(*ptr);
    end++;
    cin>>temp;
    while((temp<='z'&&temp>='a')||temp=='0'){
        if(tree[pos].value==-1){
            while(tree[pos].value==-1){
                pos++;
            }
        }
        if(temp=='0'){
            temp=-1;
        }
        ptr=new Node;
        ptr->value=temp;
        ptr->leftChild=-1;
        ptr->rightChild=-1;
        tree.push_back(*ptr);
        if(flag){
            tree[pos].rightChild=end;
        }
        else{
            tree[pos].leftChild=end;
        }
        end++;
        flag=(flag==0?1:0);
        getchar();
        if(!flag){
            pos++;
        }
        cin>>temp;
    }
    getchar();
    getchar();
    cin>>head;
    if(head=='A'){
        A(0);
    }
    else if(head=='B'){
        B(0);
    }
    else{
        C(0);
    }
}

数据很水，直接从前到后扫一遍找父节点就行了，为什么AC的人那么少……
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

int i, j, k, l, m, n, lson[2015], rson[2015];
char a[2015], ch;

void xxbl(int x){
if (a[x] != '0') putchar(a[x]);
if (lson[x] != -1) xxbl(lson[x]);
if (rson[x] != -1) xxbl(rson[x]);

}

void zxbl(int x){
if (lson[x] != -1) zxbl(lson[x]);
if (a[x] != '0') putchar(a[x]);
if (rson[x] != -1) zxbl(rson[x]);
}

void hxbl(int x){
if (lson[x] != -1) hxbl(lson[x]);
if (rson[x] != -1) hxbl(rson[x]);
if (a[x] != '0') putchar(a[x]);
}

int main(){
scanf("%c", &a[1]); i = 2;
getchar();
for(j = 1; j <= 2015; j ++){lson[j] = -1; rson[j] = -1;}
while(true){
ch = getchar();
if (ch == '-') break;
if (ch != ' '){
a[i] = ch;
for(j = 1; j <= i; j ++){
if (lson[j] == -1 && a[j] != '0') {
lson[j] = i;
break;
}
else if (rson[j] == -1 && a[j] != '0'){
rson[j] = i;
break;
}
}

i ++;
}
}
getchar(); getchar();
ch = getchar();
if (ch == 'A') xxbl(1);
if (ch == 'B') zxbl(1);
if (ch == 'C') hxbl(1);
return 0;
}

模拟水题
var lc,rc:array[1..20000] of longint;
a:array[1..20000] of char;
h,t:longint;
c:char;
procedure sea1(r:longint);
begin
if a[r]='0' then exit;
write(a[r]);
if lc[r]<>0 then sea1(lc[r]);
if rc[r]<>0 then sea1(rc[r]);
end;
procedure sea2(r:longint);
begin
if a[r]='0' then exit;
if lc[r]<>0 then sea2(lc[r]);
write(a[r]);
if rc[r]<>0 then sea2(rc[r]);
end;
procedure sea3(r:longint);
begin
if a[r]='0' then exit;
if lc[r]<>0 then sea3(lc[r]);
if rc[r]<>0 then sea3(rc[r]);
write(a[r]);
end;
begin
readln(a[1]);
h:=1;
t:=1;
read(c);
while c<>'-' do begin
if c=' ' then begin
read(c);
continue;
end;
inc(t);
a[t]:=c;
while a[h]='0' do inc(h);
if lc[h]=0 then begin
lc[h]:=t;
read(c);
continue;
end;
if rc[h]=0 then begin
rc[h]:=t;
inc(h);
read(c);
continue;
end;
end;
readln;
readln(c);
case c of
'A':sea1(1);
'B':sea2(1);
'C':sea3(1);
end;
writeln;
end.

用数组来建树，注意0和边界情况～

注意要用Ansistring T^T

数组要开大数组要开大数组要开大数组要开大数组要开大

第一个发题解 液!