PAT (Advanced Level) 1127. ZigZagging on a Tree (30) 解题报告

1127. ZigZagging on a Tree (30)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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:

812 11 20 17 1 15 8 512 20 17 11 15 8 5 1

Sample Output:

1 11 5 8 17 12 20 15

思路：根据中序后序建树，然后层次输出，注意输出的顺序，题目要求Z型输出

代码：

#include <cstdio>#include <vector>#include <iostream>#include <algorithm>#include <cmath>#include <queue>using namespace std;#define MAXN 1010000int in[MAXN], post[MAXN], l[MAXN], r[MAXN];int build(int inl, int inr, int postl, int postr){    if(inl > inr) return 0;    int rt = post[postr], p = inl, cnt;    while(rt != in[p]) p++;    cnt = p - inl;    l[rt] = build(inl, p-1, postl, postl + cnt - 1);    r[rt] = build(p+1, inr, postl + cnt, postr-1);    return rt;}void print(int rt){    queue<int> Q;    int level = 0;    Q.push(rt);    printf("%d", rt);    while(!Q.empty())    {        vector<int> V;        while(!Q.empty())        {            int now = Q.front();            Q.pop();            if(l[now]) V.push_back(l[now]);            if(r[now]) V.push_back(r[now]);        }        for(int i = 0; i < V.size(); i++)            Q.push(V[i]);        if(level % 2 == 0)            for(int i = 0; i < V.size(); i++)                printf(" %d", V[i]);        else            for(int i = V.size() - 1; i >= 0; i--)                printf(" %d", V[i]);        level ++;    }    printf("\n");}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]);    build(0, n-1, 0, n-1);    print(post[n-1]);    return 0;}