【PAT】1086. Tree Traversals Again (25)

import java.util.Scanner;class TreeNode {int val;TreeNode left;TreeNode right;TreeNode(int val) {this.val = val;}}/** *  * 结题思路：根据输入，构建二叉树。每当一行读入为Push，就在new一个节点给父节点对应的儿子。每当一行读入为Pop ，就返回null给父节点对应的儿子。 */public class Main {static Scanner input = new Scanner(System.in);private static int n;private static int count = 0;private static StringBuffer resultBuffer = new StringBuffer();public static void postorder(TreeNode root) {if (root != null) {postorder(root.left);postorder(root.right);resultBuffer.append(root.val + " ");}}//创建一棵树，然后返回它的根节点public static TreeNode buildTree() {TreeNode node = null;if (count < 2 * n) {String line = input.nextLine().trim();if (line.startsWith("Push")) {int value = Integer.parseInt(line.split(" ")[1]);node = new TreeNode(value);count++;} else {count++;return null;}node.left = buildTree();node.right = buildTree();}return node;}public static void main(String[] args) {//当nextint和nextlne一起用的时候会出错，全部都用nextlinen = Integer.parseInt(input.nextLine().trim());TreeNode rootNode = buildTree();postorder(rootNode);System.out.println(resultBuffer.toString().trim());}}