PAT--1127. ZigZagging on a Tree

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

题解

先由中序、后序序列构建出二叉树，再求出层序遍历的序列，每一层的序列是一个数组，输出的时候，第0层之外的偶数层逆转就ok啦。

#include <iostream>#include <string>#include <algorithm>#include <vector>#include <cstring>#include <cstdio>#include <cmath>#include <queue>using namespace std;const int maxn = 50 + 5;int n;int in[maxn], post[maxn];struct TreeNode{    int key;    TreeNode* left;    TreeNode* right;    TreeNode(int k):key(k), left(NULL), right(NULL){}};TreeNode* build(int* in, int* post, int n){    if(n == 0) return NULL;    TreeNode* node = new TreeNode(post[n - 1]);    int pos = 0;    for(; pos < n && in[pos] != post[n - 1]; ++pos) ;    int left = pos, right = n - pos - 1;    node->left = build(in, post, left);    node->right = build(in + left + 1, post + left, right);    return node;}void zigzagging(TreeNode* root){    queue<TreeNode*> Q;    vector<vector<int>> ans;    vector<int> tmp;    Q.push(root);    Q.push(NULL);    while(!Q.empty()){        TreeNode* p = Q.front(); Q.pop();        if(p == NULL){            ans.push_back(tmp);            tmp.clear();            if(Q.size() > 0) Q.push(NULL);        }else{            tmp.push_back(p->key);            if(p->left)  Q.push(p->left);            if(p->right) Q.push(p->right);        }    }    for(int i = 0; i < ans.size(); ++i){        if(i && i % 2 == 0) reverse(ans[i].begin(), ans[i].end());    }    for(int i = 0; i < ans.size(); ++i){        for(int j = 0; j < ans[i].size(); ++j){            if(i || j) cout << " ";            cout << ans[i][j];        }    }}int main(){#ifndef ONLINE_JUDGEfreopen("data.in", "r", stdin);#endif // ONLINE_JUDGE    cin >> n;    for(int i = 0; i < n; ++i) cin >> in[i];    for(int i = 0; i < n; ++i) cin >> post[i];    TreeNode* root = build(in, post, n);    zigzagging(root);    return 0;}