Suppose that all the keys in a binary tree are distinct positive integers. Given the postorder and inorder traversal sequences, you are supposed to output the level order traversal sequence of the corresponding binary tree.
Input Specification:
Each input file contains one test case. For each case, the first line gives a positive integer N (≤), the total number of nodes in the binary tree. The second line gives the postorder sequence and the third line gives the inorder sequence. All the numbers in a line are separated by a space.
Output Specification:
For each test case, print in one line the level order traversal sequence of the corresponding binary tree. 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:
7
2 3 1 5 7 6 4
1 2 3 4 5 6 7
Sample Output:
4 1 6 3 5 7 2分析：根据后序遍历和中序遍历 输出层序遍历我的做法是利用后序遍历和中序遍历建树 再层序遍历输出利用两个变量来记录我们需要建子树的范围 另一个变量记录每层的根节点 1 #define _CRT_SECURE_NO_WARNINGS
2 #include<iostream>
3 #include<vector>
4 #include<queue>
5 #include<stack>
6 #include<algorithm>
7 using namespace std;
8 typedef struct BtNode* Bt;
9 struct BtNode
10 {
11 int Num;
12 Bt RT;
13 Bt LT;
14 };
15 vector<vector<int> > Order(2);
16 Bt BuildTree(Bt T,int i2,int j2,int j1)
17 {
18 if (i2 >= j2)
19 return T;
20 int i = i2;
21 for (; i < j2; i++)
22 {
23 if (Order[1][i] == Order[0][j1 - 1])
24 break;
25 }
26 if (!T)
27 {
28 T = new BtNode();
29 T->LT = NULL;
30 T->RT = NULL;
31 T->Num = Order[1][i];
32 }
33 T->LT =BuildTree(T->LT,i2,i,j1-j2+i);
34 T->RT = BuildTree(T->RT,i+1,j2,j1-1);
35 return T;
36 }
37 int main()
38 {
39 int N;
40 cin >> N;
41 int num;
42 for(int i=0;i<2;i++)
43 for (int j = 0; j < N; j++)
44 {
45 cin >> num;
46 Order[i].push_back(num);
47 }
48 Bt T=NULL;
49 T = BuildTree(T,0,N,N);
50 queue<Bt> Q;
51 Q.push(T);
52 cout << T->Num;
53 while (!Q.empty())
54 {
55 if (Q.front()->LT)
56 {
57 Q.push(Q.front()->LT);
58 cout << " " << Q.front()->LT->Num;
59 }
60 if (Q.front()->RT)
61 {
62 Q.push(Q.front()->RT);
63 cout << " " << Q.front()->RT->Num;
64 }
65 Q.pop();
66 }
67 return 0;
68 }
原文地址：https://www.cnblogs.com/57one/p/11929098.html