PAT A1127 ZigZagging on a Tree (30分) (不建树)

PAT甲级：A1127 ZigZagging on a Tree (30分)

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. And it is a simple standard routine to print the numbers in level-order. However, if you think the problem is too simple, then you are too naive. This time you are supposed to print the numbers in “zigzagging order” – that is, starting from the root, print the numbers level-by-level, alternating between left to right and right to left. For example, for the following tree you must output: 1 11 5 8 17 12 20 15.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (≤30), the total number of nodes in the binary tree. The second line gives the inorder sequence and the third line gives the postorder sequence. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print the zigzagging sequence of the tree in a line. 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:

8
12 11 20 17 1 15 8 5
12 20 17 11 15 8 5 1 

Sample Output:

1 11 5 8 17 12 20 15

• 题意：给一个二叉树的后序和中序遍历序列，要求输出这棵树的 “Z” 字形遍历序列。
• 分析：这题不需要建树，只用在二叉树中后序序列建树方法的基础上加上depth参数用以记录深度即可，然后通过vector数组依次记录每一层的结点，因为无论前中后遍历二叉树的访问序列都是从上至下从左至右，所以push进vector[层数]同样是按照同一层结点间的顺序的，因此该方式可行，同时在建树方法中记录下出现的最大层数。为了方便记录（其实也差不多），第一层的depth被记为0。

#include <bits/stdc++.h>
using namespace std;
const int N = 35;
int post[N], in[N], maxDepth = -1;
vector<int> ans[N];
void create(int postl, int postr, int inl, int inr, int depth) {
    if (postl > postr) return;
    maxDepth = max(maxDepth, depth);
    ans[depth].push_back(post[postr]);
    int k = inl;
    while (k <= inr && in[k] != post[postr]) k++;
    int numleft = k - inl;
    create(postl, postl + numleft - 1, inl, k - 1, depth + 1);
    create(postl + numleft, postr - 1, k + 1, inr, depth + 1);
}
int main() {
    int n;
    scanf("%d", &n);
    for (int i = 0; i < n; i++) scanf("%d", &in[i]);
    for (int i = 0; i < n; i++) scanf("%d", &post[i]);
    create(0, n - 1, 0, n - 1, 0);
    printf("%d", ans[0][0]);
    for (int i = 1; i <= maxDepth; i++) {
        if (i % 2 == 0) reverse(ans[i].begin(), ans[i].end());
        for (auto it : ans[i]) printf(" %d", it);
    }
    return 0;
}