[leetcode][Binary Tree Inorder Traversal]

最簡單的二叉樹中序遍歷題目，其實我比較好奇的輸入的「#」，如何被識別爲null。 難度係數1/5

Given a binary tree, return the inorder traversal of its nodes' values.

For example:
Given binary tree{1,#,2,3},

1
    \
     2
    /
   3

return[1,3,2].

代碼以下

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ArrayList<Integer> inorderTraversal(TreeNode root) {
        // Start typing your Java solution below
        // DO NOT write main() function
        ArrayList<Integer> list = new ArrayList();
        tarvel(root,list);
        return list;
    }
    public void tarvel(TreeNode root,ArrayList<Integer> list){
        if(root == null) return ;
        tarvel(root.left,list);
        list.add(root.val);
        tarvel(root.right,list);
    }
}

因爲子函數返回一個ArrayList<Integer>，不是void，因此最簡單的是再本身添加一個返回void的函數