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 }
原文地址：https://www.cnblogs.com/zzw1024/p/11478533.html