94. Binary Tree Inorder Traversal

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.

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inorder:left-cur-right

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/**
 * 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) {
        
        ArrayList<Integer> rst = new ArrayList<Integer>();
        helper(root, rst);
        return rst;
        
    }
    
    private void helper(TreeNode node, ArrayList<Integer> rst){
        
        if(node == null) return;
        
        // in order
        helper(node.left, rst);
        rst.add(node.val);
        helper(node.right, rst);
        
        
    }
}