1020. Tree Traversals (25)-PAT甲级真题

1020.
Tree Traversals (25)

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：给定一棵二叉树的后序遍历和中序遍历，请你输出其层序遍历的序列。这里假设键值都是互不相等的正整数。

分析：与后序中序转换为前序的代码相仿（无须构造二叉树再进行广度优先搜索~~），只不过加一个变量index，表示当前的根结点在二叉树中所对应的下标（从0开始），所以进行一次输出先序的递归的时候，就可以把根结点下标所对应的值存储在level数组中（一开始把level都置为-1表示此处没有结点），这样在递归完成后level数组中非-1的数就是按照下标排列的层序遍历的顺序~~~

如果你不知道如何将后序和中序转换为先序，请看-》http://www.liuchuo.net/archives/2090

#include <cstdio>
#include <vector>
using namespace std;
vector<int> post, in, level(100000, -1);
void pre(int root, int start, int end, int index) {
if(start > end) return ;
int i = start;
while(i < end && in[i] != post[root]) i++;
level[index] = post[root];
pre(root - 1 - end + i, start, i - 1, 2 * index + 1);
pre(root - 1, i + 1, end, 2 * index + 2);
}
int main() {
int n, cnt = 0;
scanf("%d", &n);
post.resize(n);
in.resize(n);
for(int i = 0; i < n; i++) scanf("%d", &post[i]);
for(int i = 0; i < n; i++) scanf("%d", &in[i]);
pre(n-1, 0, n-1, 0);
for(int i = 0; i < level.size(); i++) {
if(level[i] != -1 && cnt != n - 1) {
printf("%d ", level[i]);
cnt++;
} else if(level[i] != -1){
printf("%d", level[i]);
break;
}
}
return 0;
}