pat 甲级 1127. ZigZagging on a Tree (30)

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:

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：通过后序遍历中序遍历复原二叉树，再按照题目要求的顺序输出，输出要求只要对层序遍历的方式稍加改动即可。
AC代码：

#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<set>
#include<queue>
#include<map>
using namespace std;
#define INF 0x3f3f3f
#define N_MAX 30000+5
typedef long long ll;
struct Node {
  int key=INF, L, R;
  Node() {}
  Node(int key,int l,int r):key(key),L(l),R(r) {}
}node[N_MAX];


vector<int> in, post;int  n, cnt;
void dfs(int n,int l,int r) {
  if (l>r) {
    node[n].key = INF;
    return;
  }
  int root = post[cnt--];
  node[n] = Node(root, 2 * n+1, 2 * n + 2);
  int k = find(in.begin(),in.end(),root)-in.begin();
  dfs(2 * n + 2, k + 1, r);
  dfs(2 * n+1, l, k - 1);
}
int order[N_MAX];
vector<int>res; vector<int>level;
void bfs(int root) {
  queue<int>que;
  que.push(root);
  order[root] = 0;
  while (!que.empty()) {
    int p = que.front(); que.pop();
    if (node[p].key != INF) {
      res.push_back(node[p].key);
      level.push_back(order[p]);
      if (node[p].L != 0) {  order[node[p].L] = order[p] + 1;que.push(node[p].L); }
      if (node[p].R != 0) { order[node[p].R] = order[p] + 1;que.push(node[p].R);  }
    }
  }
}

int main() {
  while (scanf("%d",&n)!=EOF) {
    in.resize(n); post.resize(n);
    for (int i = 0; i < n; i++)scanf("%d",&in[i]);
    for (int i = 0; i < n; i++)scanf("%d", &post[i]);
    cnt = n - 1;
    dfs(0, 0, n - 1);
    bfs(0);
    int num = 0,orde=0;//num是每一层的计数器
    vector<int>out;
    for (int i = 0; i < res.size();) {
      out.clear();
      while (i<res.size()&&orde ==level[i]) {
        out.push_back(res[i]);
        i++;
      }
      if (orde & 1) {
        for (int j = 0; j < out.size(); j++)printf("%d%s", out[j], (i == res.size()&&j+1==out.size())? "
" : " ");
      }
      else {
        for (int j = out.size() - 1; j >= 0; j--)printf("%d%s", out[j], (i == res.size()&&j==0 )? "
" : " ");
      }
      orde++;
    }

  }
}