1119 Pre- and Post-order Traversals
Statement
Metadata
- 作者: CHEN, Yue
- 单位: 浙江大学
- 代码长度限制: 16 KB
- 时间限制: 400 ms
- 内存限制: 64 MB
Suppose that all the keys in a binary tree are distinct positive integers. A unique binary tree can be determined by a given pair of postorder and inorder traversal sequences, or preorder and inorder traversal sequences. However, if only the postorder and preorder traversal sequences are given, the corresponding tree may no longer be unique.
Now given a pair of postorder and preorder traversal sequences, you are supposed to output the corresponding inorder traversal sequence of the tree. If the tree is not unique, simply output any one of them.
Input Specification
Each input file contains one test case. For each case, the first line gives a positive integer N (
Output Specification
For each test case, first printf in a line Yes if the tree is unique, or No if not. Then print in the next line the inorder traversal sequence of the corresponding binary tree. If the solution is not unique, any answer would do. It is guaranteed that at least one solution exists. All the numbers in a line must be separated by exactly one space, and there must be no extra space at the end of the line.
Sample Input 1
Sample Output 1
Sample Input 2
Sample Output 2
Solution
#include <bits/stdc++.h>
using namespace std;
using db = double;
using ll = long long;
using ull = unsigned long long;
using pII = pair<int, int>;
using pLL = pair<ll, ll>;
#define fi first
#define se second
#define dbg(x...) \
do { \
cout << "\033[32;1m" << #x << " -> "; \
err(x); \
} while (0)
void err() {
cout << "\033[39;0m" << endl;
}
template <class T, class... Ts>
void err(const T &arg, const Ts &...args) {
cout << arg << ' ';
err(args...);
}
template <template <typename...> class T, typename t, typename... A>
void err(const T<t> &arg, const A &...args) {
for (auto &v : arg) cout << v << ' ';
err(args...);
}
const int N = 1e5 + 10, INF = 0x3f3f3f3f;
int n, Unique, a[N], b[N], id[N], c[N];
vector<int> vec;
struct E {
int son[2];
E() {
son[0] = son[1] = 0;
}
} e[N];
struct SEG {
int t[N << 2];
void build(int id, int l, int r) {
t[id] = INF;
if (l == r) {
t[id] = a[l];
return;
}
int mid = (l + r) >> 1;
build(id << 1, l, mid);
build(id << 1 | 1, mid + 1, r);
t[id] = min(t[id << 1], t[id << 1 | 1]);
}
int query(int id, int l, int r, int ql, int qr) {
if (l >= ql && r <= qr)
return t[id];
int mid = (l + r) >> 1;
int res = INF;
if (ql <= mid)
res = min(res, query(id << 1, l, mid, ql, qr));
if (qr > mid)
res = min(res, query(id << 1 | 1, mid + 1, r, ql, qr));
return res;
}
int query(int l, int r) {
return query(1, 1, n, l, r);
}
} seg;
int gao(int al, int ar, int bl, int br) {
if (al > ar || bl > br)
return 0;
int rt = a[al];
if (al == ar)
return rt;
if (a[al + 1] == b[br - 1]) {
Unique = 0;
int Min = seg.query(al + 1, ar);
if (Min < rt) {
e[rt].son[0] = gao(al + 1, ar, bl, br - 1);
} else {
e[rt].son[1] = gao(al + 1, ar, bl, br - 1);
}
} else {
int pos = id[b[br - 1]];
int lsze = pos - 1 - al;
e[rt].son[0] = gao(al + 1, pos - 1, bl, bl + lsze - 1);
e[rt].son[1] = gao(pos, ar, bl + lsze, br - 1);
}
return rt;
}
void getprint(int u) {
if (!u)
return;
getprint(e[u].son[0]);
vec.push_back(c[u]);
getprint(e[u].son[1]);
}
int main() {
scanf("%d", &n);
for (int i = 1; i <= n; ++i) {
e[i] = E();
scanf("%d", a + i);
c[i] = a[i];
}
for (int i = 1; i <= n; ++i) {
scanf("%d", b + i);
}
sort(c + 1, c + 1 + n);
for (int i = 1; i <= n; ++i) {
a[i] = lower_bound(c + 1, c + 1 + n, a[i]) - c;
b[i] = lower_bound(c + 1, c + 1 + n, b[i]) - c;
id[a[i]] = i;
}
seg.build(1, 1, n);
vec.clear();
Unique = 1;
getprint(gao(1, n, 1, n));
if (Unique)
puts("Yes");
else
puts("No");
for (int i = 0; i < n; ++i) printf("%d%c", vec[i], " \n"[i == n - 1]);
return 0;
}