标签：tput 左右 mes ++ level 位置 strong view traversal
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 (<=30), 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 7Sample Output:
4 1 6 3 5 7 2
已知后序中序求先序：
1 void pre(int root, int begin, int end) 2 { 3 if (begin > end) return; 4 cout << post[root] << " "; 5 int p; 6 for (p = begin; p <= end; p++) 7 { 8 if (in[p] == post[root]) 9 break; 10 } 11 pre(root - end + p - 1, begin, p - 1); 12 pre(root - 1, p + 1, end); 13 }
后序中最后一个元素为跟，找出跟在中序中的位置就可以确定左右子树的节点个数，然后就可以递归的去求解。
这个题是求层序遍历，开始使用的构造树然后再广度优先搜索的方法，比较繁琐。后来看柳婼的博客发现使用数组的方法构造树比较方便。
若当前根的下标为i，则左节点下标为 2 * i + 1，右节点下标为 2 * i + 2。
1 #include <iostream> 2 #include <vector> 3 #include <math.h> 4 using namespace std; 5 6 vector<int> post, in, level(100000, -1); 7 void getLevel(int root, int begin, int end, int index); 8 9 int main() 10 { 11 int N, i; 12 cin >> N; 13 post.resize(N); 14 in.resize(N); 15 for (i = 0; i < N; i++) 16 cin >> post[i]; 17 for (i = 0; i < N; i++) 18 cin >> in[i]; 19 getLevel(N - 1, 0, N - 1, 0); 20 int cnt = 0; 21 for (int i : level) 22 { 23 if (i != -1) 24 { 25 if (cnt > 0) 26 cout << " "; 27 cnt++; 28 cout << i; 29 if (cnt == N) 30 break; 31 } 32 } 33 return 0; 34 } 35 36 void getLevel(int root, int begin, int end, int index) 37 { 38 if (begin > end) return; 39 level[index] = post[root]; 40 int p; 41 for (p = begin; p <= end; p++) 42 { 43 if (in[p] == post[root]) 44 break; 45 } 46 getLevel(root - end + p - 1, begin, p - 1, 2 * index + 1); 47 getLevel(root - 1, p + 1, end, 2 * index + 2); 48 }
标签：tput 左右 mes ++ level 位置 strong view traversal
原文地址：https://www.cnblogs.com/lxc1910/p/9501289.html
