Source:
Description:
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
Keys:
- 二叉树的建立
- 二叉树的遍历
Attention:
- 开始方向弄反了-,-
Code:
1 /*
2 Data: 2019-05-31 21:17:07
3 Problem: PAT_A1127#ZigZagging on a Tree
4 AC: 33:50
5
6 题目大意：
7 假设树的键值为不同的正整数；
8 给出二叉树的中序和后序遍历，输出二叉树的“之字形”层次遍历；
9 */
10
11 #include<cstdio>
12 #include<stack>
13 #include<queue>
14 using namespace std;
15 const int M=35;
16 int in[M],post[M];
17 struct node
18 {
19 int data,layer;
20 node *lchild,*rchild;
21 };
22
23 node *Create(int postL, int postR, int inL, int inR)
24 {
25 if(postL > postR)
26 return NULL;
27 node *root = new node;
28 root->data = post[postR];
29 int k;
30 for(k=inL; k<=inR; k++)
31 if(in[k]==root->data)
32 break;
33 int numLeft = k-inL;
34 root->lchild = Create(postL, postL+numLeft-1, inL,k-1);
35 root->rchild = Create(postL+numLeft, postR-1, k+1,inR);
36 return root;
37 }
38
39 void Travel(node *root)
40 {
41 queue<node*> q;
42 stack<node*> s;
43 root->layer=1;
44 q.push(root);
45 while(!q.empty())
46 {
47 root = q.front();q.pop();
48 if(root->layer%2==0)
49 {
50 while(!s.empty())
51 {
52 printf(" %d", s.top()->data);
53 s.pop();
54 }
55 printf(" %d", root->data);
56 }
57 else{
58 if(root->layer==1)
59 printf("%d", root->data);
60 else
61 s.push(root);
62 }
63 if(root->lchild){
64 root->lchild->layer=root->layer+1;
65 q.push(root->lchild);
66 }
67 if(root->rchild){
68 root->rchild->layer=root->layer+1;
69 q.push(root->rchild);
70 }
71 }
72 while(!s.empty())
73 {
74 printf(" %d", s.top()->data);
75 s.pop();
76 }
77 }
78
79 int main()
80 {
81 #ifdef ONLINE_JUDGE
82 #else
83 freopen("Test.txt", "r", stdin);
84 #endif
85
86 int n;
87 scanf("%d", &n);
88 for(int i=0; i<n; i++)
89 scanf("%d", &in[i]);
90 for(int i=0; i<n; i++)
91 scanf("%d", &post[i]);
92 node *root = Create(0,n-1,0,n-1);
93 Travel(root);
94
95 return 0;
96 }
原文地址：https://www.cnblogs.com/blue-lin/p/10957755.html