PAT A1020

时间:2021-05-27 03:33:48

PAT A1020

标签(空格分隔): PAT


#include <cstdio>
#include <queue>
using namespace std; const int maxn = 100;
int pos[maxn], in[maxn];
int n; struct node {
int data;
node* lchild;
node* rchild;
}; node* create(int inL, int inR, int posL, int posR) {
if(posL > posR) return NULL;
node* root = new node;
root->data = pos[posR];
int k;
for(k = inL; k <= inR; k++) {
if(pos[posR] == in[k])
break;
}
int numberLeft = k - inL;
root->lchild = create(inL, inL + numberLeft - 1, posL, posL + numberLeft - 1);
root->rchild = create(inL + numberLeft + 1, inR, posL + numberLeft, posR - 1);
return root;
} int cnt = 0; //坑
void print(node* root) {
queue<node*> q;
q.push(root);
while(!q.empty()) {
node* now = q.front();
q.pop();
printf("%d", now->data);
cnt++;
if(cnt < n) printf(" ");
if(now->lchild != NULL) q.push(now->lchild);
if(now->rchild != NULL) q.push(now->rchild);
}
} int main() {
scanf("%d", &n);
for(int i = 1; i <= n; i++) {
scanf("%d", &pos[i]);
}
for(int i = 1; i <= n; i++) {
scanf("%d", &in[i]);
}
node* ans = new node;
ans = create(1, n, 1, n);
print(ans);
return 0;
}