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 7
Sample Output:
4 1 6 3 5 7 2
题目大意： n(二叉树中结点个数)
假设二叉树中的所有键都是不同的正整数，根据后序遍历和中序遍历求层次遍历
1 #include <cstdio> 2 #include <vector> 3 #include <map> 4 using namespace std; 5 vector<int> post, in; 6 map<int, int> level;
/*后序下标 中序下标 中序下标*/ 7 void pre(int root, int start, int end, int index) { 8 if(start > end) return ; 9 int i = start; 10 while(i < end && in[i] != post[root]) i++; 11 level[index] = post[root]; 12 pre(root - 1 - end + i, start, i - 1, 2 * index + 1); 13 pre(root - 1, i + 1, end, 2 * index + 2); 14 } 15 int main() { 16 int n; 17 scanf("%d", &n); 18 post.resize(n); 19 in.resize(n); 20 for(int i = 0; i < n; i++) scanf("%d", &post[i]); 21 for(int i = 0; i < n; i++) scanf("%d", &in[i]); 22 pre(n-1, 0, n-1, 0); 23 auto it = level.begin(); 24 printf("%d", it->second); 25 while(++it != level.end()) printf(" %d", it->second); 26 return 0; 27 }
解法：1、根据后续遍历和中序遍历转化为前序遍历，并给每个结点赋上index变量：因为后序的最后
一个总是根结点，令i在中序中找到该根结点，则i把中序分为两部分，左边是左子树，右边
是右子树。因为是输出先序（根左右），所以先打印出当前根结点，然后打印左子树，再打
印右子树。左子树在后序中的根结点为root – (end – i + 1)，即为当前根结点-(右子树
的个数+1)。左子树在中序中的起始点start为start，末尾end点为i – 1.右子树的根结点
为当前根结点的前一个结点root – 1，右子树的起始点start为i+1，末尾end点为end。
2、根据Map的Key按从小到大排序的特性，使用迭代器把Value值打印出来。