Leetcode-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?

Recursive：

import java.util.ArrayList;
import java.util.List;
public class Solution {
public void inorder(TreeNode root, List<Integer> list){
if( root != null ){
inorder(root.left, list);
list.add(root.val);
inorder(root.right,list);
}
}
public List<Integer> inorderTraversal(TreeNode root) {
List<Integer> list = new ArrayList<Integer>();
inorder(root, list);
return list;
}
}

Iterative:

import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;

public class Solution {
public List<Integer> inorderTraversal(TreeNode root) {
List<Integer> list = new ArrayList<Integer>();
if( root == null )
return list;
LinkedList<TreeNode> stack = new LinkedList<TreeNode>();
while( root != null || !stack.isEmpty() ){
if( root != null ){
stack.push(root);
root = root.left;
}else{
root = stack.pop();
list.add(root.val);
root = root.right;
}
}
return list;
}
}