Leetcode: Binary Tree Inorder Transversal

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 方法：

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

Iterative method: 参考了一下网上的思路，其实就是用一个栈来模拟递归的过程。所以算法时间复杂度也是O(n)，空间复杂度是栈的大小O(logn)。

 /**
  * Definition for a binary tree node.
  * public class TreeNode {
  *     int val;
  *     TreeNode left;
  *     TreeNode right;
  *     TreeNode(int x) { val = x; }
  * }
  */
 public class Solution {
     public List<Integer> inorderTraversal(TreeNode root) {
         List<Integer> res = new ArrayList<>();
         Stack<TreeNode> stack = new Stack<>();
         TreeNode p = root;
         while (p!=null || !stack.isEmpty()) {
             if (p != null) {
                 stack.push(p);
                 p = p.left;
             }
             else {
                 TreeNode node = stack.pop();
                 res.add(node.val);
                 p = node.right;
             }
         }
         return res;
     }
 }