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;
    }
}