A1127. 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

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<algorithm>
 4 #include<queue>
 5 #include<vector>
 6 using namespace std;
 7 typedef struct NODE{
 8     struct NODE* lchild, *rchild;
 9     int data;
10     int lev;
11 }node;
12 int post[31], in[31], N;
13 node* create(int inL, int inR, int postL, int postR, int level){
14     if(inL > inR){
15         return NULL;
16     }
17     node* root = new node;
18     root->data = post[postR];
19     root->lev = level;
20     int mid;
21     for(mid = inL; mid <= inR; mid++){
22         if(in[mid] == root->data)
23             break;
24     }
25     int len = mid - inL;
26     root->lchild = create(inL, mid - 1, postL, postL + len - 1, level + 1);
27     root->rchild = create(mid + 1, inR, postL + len, postR - 1, level + 1);
28     return root;
29 }
30 void levelOrder(node* root){
31     queue<node*> Q;
32     Q.push(root);
33     int nowLev = root->lev;
34     int reverTag = 1;
35     int cnt = 0;
36     vector<node*> vec;
37     while(Q.empty() == false){
38         node* temp = Q.front();
39         Q.pop();
40         if(temp->lev == nowLev){
41             vec.push_back(temp);
42         }else{
43             nowLev = temp->lev;
44             if(reverTag == 1){
45                 reverTag = 0;
46                 for(int i = vec.size() - 1; i>= 0; i--){
47                     cnt++;
48                     printf("%d ", vec[i]->data);
49                 }
50             }else{
51                 reverTag = 1;
52                 for(int i = 0; i < vec.size(); i++){
53                     cnt++;
54                     printf("%d ", vec[i]->data);
55                 }
56             }
57             vec.clear();
58             vec.push_back(temp);
59         }
60         if(temp->lchild != NULL)
61             Q.push(temp->lchild);
62         if(temp->rchild != NULL)
63             Q.push(temp->rchild);
64     }
65     if(reverTag == 0){
66         for(int i = 0; i < vec.size(); i++){
67             cnt++;
68             if(cnt == N)
69                 printf("%d", vec[i]->data);
70             else printf("%d ", vec[i]->data);
71         }
72     }else{
73         for(int i = vec.size() - 1; i >= 0; i--){
74             cnt++;
75             if(cnt == N)
76                 printf("%d", vec[i]->data);
77             else printf("%d ", vec[i]->data);
78         }
79     }
80 }
81 int main(){
82     scanf("%d", &N);
83     for(int i = 1; i <= N; i++){
84         scanf("%d", &in[i]);
85     }
86     for(int i = 1; i <= N; i++){
87         scanf("%d",&post[i]);
88     }
89     node* root = create(1, N, 1, N, 1);
90     levelOrder(root);
91     cin >> N;
92     return 0;
93 }
View Code

 

总结:

1、题意:中序后序建树,再用层序输出,输出时一行逆序一行正序。

2、可以在建树时顺便将层次信息也加入节点。在层序遍历时使用一个vector,在当前层数相同时,仅仅把本该访问的节点存入vector,当层数发生改变时,输出vector内所有元素(设置一个计数器,如果上次是正序输出,则本次逆序输出); 或者正常层序遍历,再多用一个栈+一个队列,当需要正序输出该层时,节点入队列,需要逆序则节点入栈,层数改变时输出栈中或队列中全部节点,并清空。

 

posted @ 2018-03-04 20:02  ZHUQW  阅读(331)  评论(0编辑  收藏  举报