【题目链接】:http://hihocoder.com/problemset/problem/1049

【题意】

【题解】



前序遍历的第一个节点;

肯定是整颗树的头结点;

然后在中序遍历中;

得到这个树的左子树和右子树;

然后再分别得到左子树和右子树的前序遍历;

递归处理就好;



【Number Of WA】



0



【完整代码】

#include <bits/stdc++.h>

using namespace std;

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define LL long long

#define rep1(i,a,b) for (int i = a;i <= b;i++)

#define rep2(i,a,b) for (int i = a;i >= b;i--)

#define mp make_pair

#define pb push_back

#define fi first

#define se second

#define ms(x,y) memset(x,y,sizeof x)

typedef pair<int,int> pii;

typedef pair<LL,LL> pll;

const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};

const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};

const double pi = acos(-1.0);

const int N = 110;

int root,l[30],r[30];

string s1,s2;

int dfs(string s1,string s2)

{

    if (s1==""||s2=="") return 0;

    char t = s1[0];

    int x = t-'A'+1;

    int pos = s2.find(t,0);

    int llen = pos,rlen = s2.size()-llen-1;

    l[x] = dfs(s1.substr(1,llen),s2.substr(0,llen));

    r[x] = dfs(s1.substr(1+llen,rlen),s2.substr(pos+1,rlen));

    return x;

}

void dfs(int x)

{

    if (x==0) return;

    dfs(l[x]);dfs(r[x]);

    char t = x+'A'-1;

    cout << t;

}

int main()

{

    //freopen("F:\\rush.txt","r",stdin);

    ios::sync_with_stdio(false),cin.tie(0);//scanf,puts,printf not use

    cin >> s1 >> s2;

    root = dfs(s1,s2);

    dfs(root);

    cout << endl;

    return 0;

}