PAT_A1127#ZigZagging on a Tree

Posted 2021-12-14 blue-lin

Source:

PAT A1127 ZigZagging on a Tree (30 分)

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