题目描述：
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
解题思路：
根据中序和后序创建二叉树，最后蛇形层次遍历（遍历一层就开始反转一次，Java中使用了LinkedList进行存储处理）
代码中有根据中序和后序创建二叉树的模板，有层次遍历的模板；
先序和中序创建二叉树的模板参考先序和中序创建二叉树
代码：
package text_9_25; import java.util.ArrayList; import java.util.LinkedList; import java.util.Queue; import java.util.Scanner; public class ZigZaggingOnATree { public static void main(String[] args) { // TODO Auto-generated method stub Scanner scanner = new Scanner(System.in); int n = scanner.nextInt(); int in[] = new int[n]; int post[] = new int[n]; for(int i=0;i<n;i++) in[i] = scanner.nextInt(); for(int i=0;i<n;i++) post[i] = scanner.nextInt(); //创建二叉树 Node root = BuildTree(in, 0, n-1, post, 0, n-1); //蛇形层次遍历 PrintLevelOrder(root); } public static Node BuildTree(int in[],int ibegin,int iend,int post[],int pbegin,int pend) { if(ibegin>iend||pbegin>pend) return null; Node root = new Node(post[pend]); for(int i=ibegin;i<=iend;i++) { if(post[pend]==in[i]) { root.left = BuildTree(in, ibegin, i-1, post, pbegin, pbegin-ibegin+i-1); root.right = BuildTree(in, i+1, iend, post, pbegin-ibegin+i, pend-1); } } return root; } public static void PrintLevelOrder(Node root) { boolean reversed = false;//反转标志 Queue<Node> queue = new LinkedList<>(); queue.add(root); int flag=1;//用于输出格式控制 while(!queue.isEmpty()) { int size = queue.size();//当前层的节点数 LinkedList<Integer> link = new LinkedList<>();//保存当前层结果，LinkedList可以前后插入 for(int i=0;i<size;i++) { //对应每一层 Node temp = queue.poll(); if(reversed) link.add(temp.data); else link.addFirst(temp.data); if(temp.left!=null) queue.add(temp.left); if(temp.right!=null) queue.add(temp.right); } // System.out.println(size); for(int j=0;j<link.size();j++) { if(flag==1) { System.out.print(link.get(j)); flag=0; } else System.out.print(" "+link.get(j)); } reversed =!reversed; //每次反转 } } } class Node{ //树节点类 public int data; public Node left; public Node right; public Node(int data){ this.data = data; left = right = null; } }
