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
题目大意：通过中序遍历和后序遍历构建一棵树， 并交替的输出层序遍历；思路：dfs()构建这棵树， levelOrder()交替的层序遍历这棵树； 用两个二维数组来记录层序遍历， level1记录从右到左边的遍历， level2用来记录从左到右的遍历 设置两个计数cnt1，cnt2来记录level1和了level2的位置注意点：应该对建立树的数组进行初始化，初始化为负数， 若不初始化， 默认值是1， 然而数组有下标是0，会导致第一位数不能遍历，导致错误 1 #include<iostream>
2 #include<vector>
3 #include<queue>
4 using namespace std;
5 vector<int> in(30), post(30), level1[30], level2[30];
6 int v[30][2], root;
7 void dfs(int &index, int inl, int inr, int postl, int postr){
8 int temp=post[postr], i=inl;
9 if(inl>inr) return;
10 index = postr;
11 while(i<=inr && in[i]!=temp) i++;
12 dfs(v[index][0], inl, i-1, postl, postl+(i-inl)-1);
13 dfs(v[index][1], i+1, inr, postl+i-inl, postr-1);
14 }
15
16 void levelOrder(){
17 queue<int> q1, q2;
18 q1.push(root);
19 int cnt1=0, cnt2=0;
20 while(q1.size() || q2.size()){
21 int temp, flag1=0, flag2=0;
22 while(q1.size()){
23 temp=q1.front();
24 q1.pop();
25 level1[cnt1].push_back(post[temp]);
26 if(v[temp][0]!=-1) q2.push(v[temp][0]);
27 if(v[temp][1]!=-1) q2.push(v[temp][1]);
28 flag1=1;
29 }
30 if(flag1) cnt1++;
31 while(q2.size()){
32 temp=q2.front();
33 q2.pop();
34 level2[cnt2].push_back(post[temp]);
35 if(v[temp][0]!=-1) q1.push(v[temp][0]);
36 if(v[temp][1]!=-1) q1.push(v[temp][1]);
37 flag2=1;
38 }
39 if(flag2) cnt2++;
40 }
41 }
42 int main(){
43 int n, i, j;
44 cin>>n;
45 fill(v[0], v[0]+60, -1);
46 for(i=0; i<n; i++) cin>>in[i];
47 for(i=0; i<n; i++) cin>>post[i];
48 dfs(root, 0, n-1, 0, n-1);
49 levelOrder();
50 cout<<level1[0][0];
51 for(i=1; i<n; i++)
52 if(i%2==1) for(j=0; j<level2[i/2].size(); j++) cout<<" "<<level2[i/2][j];
53 else for(j=level1[i/2].size()-1; j>=0; j--) cout<<" "<<level1[i/2][j];
54 return 0;
55 }
原文地址：https://www.cnblogs.com/mr-stn/p/9214176.html