Build a binary tree based on preorder traversal
topic description
Write a program that reads in a string of preorder traversal strings input by the user, and builds a binary tree (stored by pointers) based on the strings. For example, the following preorder traversal string: ABC##DE#G##F### where "#" represents a space, and the space character represents an empty tree. After the binary tree is established, perform an in-order traversal on the binary tree and output the traversal result.
Input description:
The input consists of 1 line of strings, the length of which does not exceed 100.
Output description:
There may be multiple sets of test data. For each set of data, output the sequence of inorder traversal after the input string is built as a binary tree. There is a space after each character. Each output takes one line.
example
Original title OJ link
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answer
import java.util.Scanner;
// 注意类名必须为 Main, 不要有任何 package xxx 信息
class TreeNode{
//创建一个树结点的类，这个类最后不要构建在Main类了里面，构建在里面是内部类
char val;
TreeNode left;
TreeNode right;
public TreeNode(char ch){
this.val = ch;
this.left = null;
this.right = null;
}
}
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
// 注意 hasNext 和 hasNextLine 的区别
while (in.hasNextLine()) {
// 注意 while 处理多个 case
String str = in.nextLine();
TreeNode root = creatTree(str);
midOrder(root);
}
}
public static int i = 0;//必须是静态的，不然在creatTree这个静态方法里面没法用
//构建二叉树
public static TreeNode creatTree(String str){
TreeNode root = null;
if(str.charAt(i) != '#'){
//只要值不是#，就证明是结点，所以先把结点创建出来
root = new TreeNode(str.charAt(i));
i++;
root.left = creatTree(str);
root.right = creatTree(str);
}
else{
i++;//本题是不需要判断i是否越界的，因为只有给定的str是合法的前驱遍历，那么i就不会无限增加，因为有递归次数做保证
}
return root;
}
//中序遍历
public static void midOrder(TreeNode root){
if(root == null){
return;
}
midOrder(root.left);
System.out.print(root.val + " ");
midOrder(root.right);
}
}