九度OJ 1385
该问题很基础，但需要注意这里只检查是否能构建二叉树，只用满足二叉树的基本性质即可。
/**
* @brief jiu du 1385
* @file 1385.cpp
* @author mianma
* @created 2014/01/20 14:16
* @edited 2014/01/21 14:11
* @type tree
* @note
*/
#include <fstream>
#include <iostream>
#include <string>
#include <cstring>
using namespace std;
#define max(a, b) ((a) > (b) ? (a) : (b))
#define min(a, b) ((a) > (b) ? (b) : (a))
#define abs(a) ((a) > 0 ? (a) : (0 - (a)))
#define CLR(vec) memset(vec, 0, sizeof(vec))
#ifdef DEBUG
ifstream in;
ofstream out;
#define CIN in
#define COUT out
#else
#define CIN cin
#define COUT cout
#endif
#define MAXN 1100
int pre_table[MAXN];
int mid_table[MAXN];
int next_table[MAXN];
int lft_table[MAXN]; /*check the num*/
int rht_table[MAXN]; /*check the num*/
struct node{
struct node *lft;
struct node *rht;
int val;
node(const int &val){
this->lft = NULL;
this->rht = NULL;
this->val = val;
}
};
struct node *root;
int n;
int search_pos(const int *vec, const int &size, const int &val){
for(int i = 0; i < size; i++)
if(vec[i] == val)
return i;
return -1;
}
int check_valid(const int *pre, const int *mid, const int &size){
CLR(lft_table);
CLR(rht_table);
for(int i = 0; i < size; i++){
++lft_table[pre[i]];
++rht_table[mid[i]];
}
for(int i = 0; i < MAXN; i++){
if(lft_table[i] > 1 || rht_table[i] > 1)
return -1;
if(lft_table[i] != rht_table[i]){
return -1;
}
}
return 0;
}
/*status mark wether build tree fail, go into with status == 1*/
struct node *build_tree(const int *pre, const int *mid, const int size, int &status){
struct node *lft, *rht, *ret;
if(0 == status)
return NULL;
if(0 == size){
return NULL;
}
#ifdef DEBUG1
COUT << "pre:\n";
for(int i = 0; i < size; i++)
COUT << pre[i] << (i == size - 1 ? "\n" : " ");
COUT << "mid:\n";
for(int i = 0; i < size; i++)
COUT << mid[i] << (i == size - 1 ? "\n" : " ");
#endif
if(1 == size){
/*we check leaf conflict here*/
if(pre[0] != mid[0]){
status = 0;
return NULL;
}else{
return (new node(pre[0]));
}
}
int val = pre[0];
/*we check node conflict here*/
int pos = search_pos(mid, size, val);
if(pos < 0){
status = 0;
return NULL;
}
int lft_size = pos;
int rht_size = size - pos - 1;
lft = build_tree(pre + 1, mid, lft_size, status);
rht = build_tree(pre + 1 + lft_size, mid + 1 + lft_size, rht_size, status);
if(0 == status)
return NULL;
//if( (lft && lft->val <= val) || (rht && rht->val <= val)){
// status = 0;
// return NULL;
//}
ret = new node(val);
ret->lft = lft;
ret->rht = rht;
return ret;
}
void visit_tree(struct node *n, int *ans, int &pos){
if(!n) return;
visit_tree(n->lft, ans, pos);
visit_tree(n->rht, ans, pos);
ans[pos++] = n->val;
}
void destory_tree(struct node *n){
if(!n) return;
destory_tree(n->lft);
destory_tree(n->rht);
delete n;
}
int main()
{
//ios_base::sync_with_stdio(0);
#ifdef DEBUG
CIN.open("./in", ios::in);
COUT.open("./out", ios::out);
#endif
while(CIN >> n && n){
for(int i = 0; i < n; i++)
CIN >> pre_table[i];
for(int i = 0; i < n; i++)
CIN >> mid_table[i];
if(check_valid(pre_table, mid_table, n) < 0){
COUT << "No\n";
continue;
}
int status = 1;
int pos = 0;
root = build_tree(pre_table, mid_table, n, status);
visit_tree(root, next_table, pos);
if(status){
for(int i = 0; i < n; i++)
COUT << next_table[i] << " ";
COUT << "\n";
}else{
COUT << "No\n";
}
destory_tree(root);
}
return 0;
}