# PAT 1127. ZigZagging on a Tree (30)-甲级

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

2.进行广度优先搜索，将树从根结点开始所有结点层序遍历，保存在result二维数组中，比如：result[i]保存第i层所有结点的序列～
3.进行z字型输出，根据当前层号的奇偶性分别从左往右、从右往左遍历输出～

1. dfs：因为post(后序)是按照左、右、根的顺序遍历的，所以从右往左，最右边的肯定是根结点～所以postRight是当前子树的根结点的下标，将它的赋值给index，并继续dfs tree[index][0]和tree[index][1]～

2.广度优先搜索，采用队列q，q中保存的是node结点，node.index表示当前节点在post中的下标，node.depth表示当前结点在树中的层数～
3.当 i % 2 == 0的时候倒序输出，否则正序输出～

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