Binary Tree Inorder Traversal

Link: http://oj.leetcode.com/problems/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.

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}".

Recusion Version:

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    ArrayList<Integer> nodes = new ArrayList<Integer>();
    public ArrayList<Integer> inorderTraversal(TreeNode root) {
        inorderTraversal_recursion(root);
        return nodes;
    }
    public void inorderTraversal_recursion(TreeNode root){
        if(root==null)
            return;
        inorderTraversal_recursion(root.left);
        nodes.add(root.val);
        inorderTraversal_recursion(root.right);
    }
    
}

Iterative version:

/**
 * 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) {
        // the nodes is used to store the result
        ArrayList<Integer> nodes = new ArrayList<Integer>();
        if (root == null)
            return nodes;
        Stack<TreeNode> stack = new Stack<TreeNode>();
        TreeNode node = root;
        // note in the begin of the loop, stack is empty,
        // so we need node != null to make sure the while-loop
        // can be executed.
        while (node != null || !stack.isEmpty()) {
            // if the node is not null,
            // we should get the left child
            if (node != null) {
                stack.push(node);
                node = node.left;
            }
            // if the node is null
            // i.e. reach the situation that the parent
            // don't have left child.
            // we should pop the top of the stack,
            // and do the operation on the right child.
            else {
                node = stack.pop();
                nodes.add(node.val);
                node = node.right;
            }
        }
        return nodes;
    }
}