学步园

Given binary tree {1,#,2,3},

   1
    \
     2
    /
   3

return [1,3,2].

Can you do it without recursion?

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */
public class Solution {
    /**
     * @param root: The root of binary tree.
     * @return: Inorder in ArrayList which contains node values.
     */
    public static ArrayList<Integer> inorderTraversal(TreeNode root) {
		ArrayList<Integer> tree = new ArrayList<Integer>();
		Stack<TreeNode> s = new Stack<TreeNode>();

		while (root != null) {
			if (root.left == null) {
				tree.add(root.val);
				if (root.right != null) {
					TreeNode r = root.right;
					root.right = null;
					root = r;
				} else if (!s.isEmpty())
					root = s.pop();
				else
					root = null;
			} else if (root.left != null) {
				TreeNode l = root.left;
				root.left = null;
				s.push(root);
				root = l;
			}
		}
		return tree;
	}
}