真二叉樹重構(Proper Rebuild)

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Description

In general, given the preorder traversal sequence and postorder traversal sequence of a binary tree, we cannot determine the binary tree.

Figure 1

In Figure 1 for example, although they are two different binary tree, their preorder traversal sequence and postorder traversal sequence are both of the same.

But for one proper binary tree, in which each internal node has two sons, we can uniquely determine it through its given preorder traversal sequence and postorder traversal sequence.

Label n nodes in one binary tree using the integers in [1, n], we would like to output the inorder traversal sequence of a binary tree through its preorder and postorder traversal sequence.

Input

The 1st line is an integer n, i.e., the number of nodes in one given binary tree,

The 2nd and 3rd lines are the given preorder and postorder traversal sequence respectively.

Output

The inorder traversal sequence of the given binary tree in one line.

Example

Input

5
1 2 4 5 3
4 5 2 3 1

Output

4 2 5 1 3

Restrictions

For 95% of the estimation, 1 <= n <= 1,000,00

For 100% of the estimation, 1 <= n <= 4,000,000

The input sequence is a permutation of {1,2...n}, corresponding to a legal binary tree.

Time: 2 sec

Memory: 256 MB

Hints

Figure 2

In Figure 2, observe the positions of the left and right children in preorder and postorder traversal sequence.

1. 原理與要點：利用樹的先序與后序的規律，遞歸建樹，然后dfs輸出樹的中序遍歷。可以發現先序的第一個節點是根節點，后序的最后一個節點是根節點，利用這個規律確定左右子樹，一直遞歸下去，就可以確定每一個節點
2. 遇到的問題：無
3. 時間和空間復雜度：時間復雜度\(O(nlogn)\)，空間復雜度\(O(n)\)

#include <cstdlib>
#include "cstdio"

using namespace std;
const int maxn = 4e6 + 100;
const int SZ = 1 << 20;  //快速io
struct fastio {
    char inbuf[SZ];
    char outbuf[SZ];

    fastio() {
        setvbuf(stdin, inbuf, _IOFBF, SZ);
        setvbuf(stdout, outbuf, _IOFBF, SZ);
    }
} io;

typedef struct node {
    int val;
    node *l, *r;
} *tree;
tree root = NULL;

int mlr[maxn], lrm[maxn];
int n;

void build(tree &t, int s1, int e1, int s2, int e2) {

    t = (tree) malloc(sizeof(node));
    t->val = mlr[s1];
    if (s1 == e1) return;
    int now;
    for (int i = s2; i <= e2; i++) {
        if (mlr[s1 + 1] == lrm[i]) {
            now = i;
            break;
        }
    }
    build(t->l, s1 + 1, now - s2 + 1 + s1, s2, now);
    build(t->r, now - s2 + 2 + s1, e1, now + 1, e2 - 1);
}

void dfs(tree now) {
    if (now == NULL) return;
    dfs(now->l);
    printf("%d ", now->val);
    dfs(now->r);
}

int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
        scanf("%d", &mlr[i]);
    }
    for (int i = 1; i <= n; i++) {
        scanf("%d", &lrm[i]);
    }
    build(root, 1, n, 1, n);
    dfs(root);
    return 0;
}