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