ZigZagging on a Tree

Origin problem

Suppose that all the keys in a binary tree are distinct positive integers. A unique binary tree can be determined by a given pair of postorder and inorder traversal sequences. And it is a simple standard routine to print the numbers in level-order. However, if you think the problem is too simple, then you are too naive. This time you are supposed to print the numbers in “zigzagging order” – that is, starting from the root, print the numbers level-by-level, alternating between left to right and right to left. For example, for the following tree you must output: 1 11 5 8 17 12 20 15.

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 inorder sequence and the third line gives the postorder sequence. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print the zigzagging sequence of the tree in a line. 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:

8
12 11 20 17 1 15 8 5
12 20 17 11 15 8 5 1

Sample Output:

1 11 5 8 17 12 20 15

思路

这一题最重要的一点是，理解inorder和postorder的输出顺序：

可以看到，通过postorder最后一个数即可得到该树的根，进而通过inorder中的根得到树的左右分支：

例如：sample中postorder最后输出为1，即整个树的根为1，再通过inorder中1的位置可以得到，左分支为12 11 20 17构成，右分支为15 8 5构成。

在这一基础上,，可以考虑分而治之的做法，即每次获得一个根，然后递归进入左支和右支即可，细节如下：

• 树每层的数据可以使用一个数组进行存储，为了方便~~(懒)~~，我用的二维数组作为栈以及一组int指针，能顺序存数据，也能记录个数。
• 需要使用到4个指针，分别指向inorder和postorder的开始和结束。
• 对于inorder，获得其根的index后就很简单，左右数据分别为左右分支。
• 对于postorder，通过inorder中获得的左支元素个数，即可获得postorder左支的index，进而获得右支的index
  • 这里要
• 在获得上述index后，递归即可，应该注意的是，先递归左侧的，在递归右侧的。
• 最后把数组中数据按要求输出即可。