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：

 1 /**
 2  * Definition for binary tree
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public List<Integer> inorderTraversal(TreeNode root) {
12         ArrayList<Integer> res = new ArrayList<Integer>();
13         if (root == null) return res;
14         helper(root, res);
15         return res;
16     }
17     
18     public void helper(TreeNode root, ArrayList<Integer> res) {
19         if (root == null) {
20             return;
21         }
22         helper(root.left, res);
23         res.add(root.val);
24         helper(root.right, res);
25     }
26 }

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)。

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public List<Integer> inorderTraversal(TreeNode root) {
12         List<Integer> res = new ArrayList<>();
13         Stack<TreeNode> stack = new Stack<>();
14         TreeNode p = root;
15         while (p!=null || !stack.isEmpty()) {
16             if (p != null) {
17                 stack.push(p);
18                 p = p.left;
19             }
20             else {
21                 TreeNode node = stack.pop();
22                 res.add(node.val);
23                 p = node.right;
24             }
25         }
26         return res;
27     }
28 }