binary-tree-inorder-traversal Java code

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].

Note: Recursive solution is trivial, could you do it iteratively?

confused what”{1,#,2,3}”means? > read more on how binary tree is serialized on OJ.

OJ’s Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where ‘#’ signifies a path terminator where no node exists below.

Here’s an example:

1

/ \

2 3

/

4

\

5

The above binary tree is serialized as”{1,2,3,#,#,4,#,#,5}”.

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
import java.util.ArrayList;
public class Solution {
    public ArrayList<Integer> inorderTraversal(TreeNode root) {
        ArrayList<Integer>  res=new ArrayList<Integer>();
        if(root==null)return res;
        inorder(root,res);
        return res;
    }
     private static void inorder(TreeNode root, ArrayList<Integer> list){
        if(root != null){
            inorder(root.left,list);
            list.add(root.val);
            inorder(root.right,list);
        }
    }
}