1127 ZigZagging on a Tree （30 分）

最新推荐文章于 2021-03-09 11:40:09 发布

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原文链接：http://www.cnblogs.com/Mered1th/p/10450754.html

本文介绍了一个二叉树的ZigZag遍历算法，通过中序和后序遍历序列构建二叉树，然后以ZigZag顺序输出节点。算法首先使用层序遍历将节点存入容器，再根据层级奇偶性逆序或正序输出节点值。

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1127 ZigZagging on a Tree （30 分）

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 (

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

题意：中序和后序建树，然后按zigzagging order输出。

分析：层序遍历的时候将节点输出到容器中，最后输出的时候根据奇数还是偶数来输出结点

 1 /**
 2 * Copyright(c)
 3 * All rights reserved.
 4 * Author : Mered1th
 5 * Date : 2019-02-28-14.24.50
 6 * Description : A1127
 7 */
 8 #include<cstdio>
 9 #include<cstring>
10 #include<iostream>
11 #include<cmath>
12 #include<algorithm>
13 #include<string>
14 #include<unordered_set>
15 #include<map>
16 #include<vector>
17 #include<set>
18 #include<queue>
19 using namespace std;
20 const int maxn=31;
21 int n,post[maxn],in[maxn];
22 vector<int> ans[maxn];
23 struct Node{
24     int val,layer;
25     Node* lchild;
26     Node* rchild;
27     Node(int _val,int _layer){
28         val=_val;
29         lchild=NULL;
30         rchild=NULL;
31         layer=_layer;
32     }
33 };
34 int maxlayer=1;
35 Node* create(int inL,int inR,int postL,int postR,int layer){
36     if(inL>inR) return NULL;
37     if(layer>maxlayer) maxlayer=layer;
38     int rootVal=post[postR];
39     Node* root=new Node(rootVal,layer);
40     int k;
41     for(int i=inL;i<=inR;i++){
42         if(post[postR]==in[i]){
43             k=i;
44             break;
45         }
46     }
47     int numLeft=k-inL;
48     root->lchild=create(inL,k-1,postL,postL+numLeft-1,layer+1);
49     root->rchild=create(k+1,inR,postL+numLeft,postR-1,layer+1);
50     return root;
51 }
52 
53 void BFS(Node* root){
54     queue<Node*> q;
55     q.push(root);
56     while(!q.empty()){
57         Node* now=q.front();
58         q.pop();
59         ans[now->layer].push_back(now->val);
60         if(now->lchild!=NULL) q.push(now->lchild);
61         if(now->rchild!=NULL) q.push(now->rchild);
62     }
63 }
64 
65 int main(){
66 #ifdef ONLINE_JUDGE
67 #else
68     freopen("1.txt", "r", stdin);
69 #endif
70     scanf("%d",&n);
71     for(int i=0;i<n;i++){
72         scanf("%d",&in[i]);
73     }
74     for(int i=0;i<n;i++){
75         scanf("%d",&post[i]);
76     }
77     Node* root=create(0,n-1,0,n-1,1);
78     BFS(root);
79     for(int i=1;i<=maxlayer;i++){
80         if(i==1){
81             printf("%d",ans[i][0]);
82             continue;
83         }
84         if(i%2==0){
85             for(int j=0;j<ans[i].size();j++){
86                 printf(" %d",ans[i][j]);
87             }
88         }
89         else{
90             for(int j=ans[i].size()-1;j>=0;j--){
91                 printf(" %d",ans[i][j]);
92             }
93         }
94     }
95     return 0;
96 }

转载于:https://www.cnblogs.com/Mered1th/p/10450754.html