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

题目大意:给出一个树的中序和后序遍历结果,求它的Z字型层序遍历,也就是偶数层从右往左,奇数层从左往右遍历~
分析:分为3步:1.根据中序和后序建树 保存在tree二维数组中,比如:tree[i][0] = val表示post[i]的左孩子是post[val],tree[i][1] = val表示post[i]的右孩子是post[val]~
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]~
根据post[postRight]的结点在in里面的下标位置i,可以得到i的左边是左子树,即inLeft 到 i – 1,右边是右子树:i + 1 到 inRight。而对于post来说,根据左子树的结点个数i – inLeft可以得到[postLeft, postLeft + (i – inLeft) – 1]是post中左子树的范围,[postLeft + (i – inLeft), postRight – 1]是post中右子树的范围~
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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