题目描述：
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
题目思路：
dfs从根节点递归左右子树，记录每个结点层次遍历的对应坐标，最后遍历输出即可。
题目代码：
#include <cstdio>
#include <cstring>
using namespace std;int n;
int pod[35];
int iod[35];
int lod[10000];void dfs(int p1, int p2, int q1, int q2, int od){if(p1 > p2 || q1 > q2) return ;int i = p1;while(iod[i] != pod[q2]) i++;lod[od] = pod[q2]; dfs(p1, i-1, q1, q1+i-1-p1, 2*od);dfs(i+1, p2, q1+i-p1, q2-1, 2*od+1);
}int main(){scanf("%d",&n);memset(lod,0,sizeof(lod));for(int i = 0; i < n; i++){scanf("%d",&pod[i]);}for(int i = 0; i < n; i++){scanf("%d",&iod[i]);}dfs(0, n-1, 0, n-1, 1);int flag = 1;for(int i = 1; i < 10000; i++){if(flag && lod[i] != 0){printf("%d",lod[i]);flag = 0;} else if(!flag && lod[i]){printf(" %d",lod[i]);}}return 0;
}