题解 | #二叉树遍历#
二叉树遍历
https://www.nowcoder.com/practice/4b91205483694f449f94c179883c1fef
import java.util.Scanner;
// 注意类名必须为 Main, 不要有任何 package xxx 信息
public class Main {
static class TreeNode {
public char val;
public TreeNode left;
public TreeNode right;
TreeNode(char ch) {
this.val = ch;
}
}
static int count = 0;
public static TreeNode createTree(String str) {
char ch = str.charAt(count++);
if(count == str.length() || ch == '#') {
return null;
}
TreeNode root = new TreeNode(ch);
root.left = createTree(str);
root.right = createTree(str);
return root;
}
public static void midOrderTraverse(TreeNode root) {
if(root == null) {
return;
}
midOrderTraverse(root.left);
System.out.print(root.val+" ");
midOrderTraverse(root.right);
}
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
String str = sc.nextLine();
TreeNode root = createTree(str);
midOrderTraverse(root);
}
}
设置一个控制变量控制字符串的读取位置,按照前序遍历的顺序,通过前序遍历的递归样式创建树,如果是# 就直接返回null,如果不是就创建节点,再对左子树和右子树进行创建和接受引用,然后返回当前节点