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}".

 

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    ArrayList<Integer> result = null;
    public ArrayList<Integer> inorderTraversal(TreeNode root) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        result = new ArrayList<Integer>();
        inorder(root);
        return result;
    }
    public void inorder(TreeNode root){
        if(root == null) return;
        inorder(root.left);
        result.add(root.val);
        inorder(root.right);
    }
}

//1.build threaded binary tree
public List<Integer> inorderTraversal(TreeNode root) {
    // write your code here
    TreeNode cur = null, pre = null;
    List<Integer> result = new ArrayList<Integer>();
    if(root == null) return result;
    cur = root;
    while(cur != null){
        if(cur.left == null){
            result.add(cur.val);
            cur = cur.right;
        }else{
            pre = cur.left;
            while(pre.right != null && pre.right != cur){
                pre = pre.right;
            }
            if(pre.right == null){//first time: build threaded binary tree
                pre.right = cur;
                cur = cur.left;
            }else{//second time: recover tree and output val
                result.add(cur.val);
                pre.right = null;
                cur = cur.right;
            }
            
        }
    }
    return result;
}

//2. stack
public List<Integer> inorderTraversal(TreeNode root) {
    List<Integer> result = new ArrayList<Integer>();
    if(root == null) return result;
    LinkedList<TreeNode> stack = new LinkedList<TreeNode>();
    boolean flg = true;
    while(flg){
        if(root != null){
            stack.push(root);
            root = root.left;
        }else{
            if(stack.isEmpty()) flg = false;
            else{
                root = stack.pop();
                result.add(root.val);
                root = root.right;
            }
        }
    }
    return result;
}