PAT甲级——A1127 ZigZagging on a Tree30

Posted 2022-10-06 zzw1024

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


首先通过后续遍历和中序遍历重构二叉树
然后在层序遍历的基础上，进行元素逆序

 1 #include <iostream>
 2 #include <vector>
 3 #include <queue>
 4 #include <algorithm>
 5 using namespace std;
 6 struct Node
 7 
 8     int v;
 9     Node *l, *r;
10     Node(int a = -1) :v(a), l(nullptr), r(nullptr) 
11 ;
12 int n;
13 vector<int>inOrder, posOrder, zigOrder;
14 Node* reCreatTree(int inL, int inR, int posL, int posR)
15 
16     if (posL > posR)
17         return nullptr;
18     Node* root = new Node(posOrder[posR]);
19     int k = inL;
20     while (k<=inR && inOrder[k] != posOrder[posR])++k;//找到根节点
21     int numL = k - inL;//左子树节点个数
22     root->l = reCreatTree(inL, k - 1, posL, posL + numL - 1);
23     root->r = reCreatTree(k + 1, inR, posL + numL, posR - 1);
24     return root;
25 
26 void zigOrderTravel(Node* root)
27 
28     if (root == nullptr)
29         return;
30     queue<Node*>q;
31     q.push(root);
32     zigOrder.push_back(root->v);
33     bool flag = false;
34     while (!q.empty())
35     
36         queue<Node*>temp;
37         vector<int>tt;
38         while (!q.empty())
39         
40             Node* p = q.front();
41             q.pop();
42             if (p->l != nullptr)
43             
44                 temp.push(p->l);
45                 tt.push_back(p->l->v);
46             
47             if (p->r != nullptr)
48             
49                 temp.push(p->r);
50                 tt.push_back(p->r->v);
51             
52         
53         if (flag)
54             reverse(tt.begin(), tt.end());
55         zigOrder.insert(zigOrder.end(), tt.begin(), tt.end());
56         flag = !flag;
57         q = temp;
58     
59 
60 int main()
61 
62     cin >> n;
63     inOrder.resize(n);
64     posOrder.resize(n);
65     for (int i = 0; i < n; ++i)
66         cin >> inOrder[i];
67     for (int i = 0; i < n; ++i)
68         cin >> posOrder[i];
69     Node* root = reCreatTree(0, n - 1, 0, n - 1);
70     zigOrderTravel(root);
71     for (int i = 0; i < zigOrder.size(); ++i)
72         cout << (i > 0 ? " " : "") << zigOrder[i];
73     return 0;
74