Binary Tree Inorder Traversal Solutions in Java

/// Source : https://leetcode.com/problems/binary-tree-inorder-traversal/
/// Author : liuyubobobo
/// Time   : 2017-11-17

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

// Recursive
// Time Complexity: O(n), n is the node number in the tree
// Space Complexity: O(h), h is the height of the tree
public class Solution1 {

    public List<Integer> inorderTraversal(TreeNode root) {

        ArrayList<Integer> res = new ArrayList<Integer>();
        inorderTraversal(root, res);
        return res;
    }

    private void inorderTraversal(TreeNode node, List<Integer> list){
        if(node != null){
            inorderTraversal(node.left, list);
            list.add(node.val);
            inorderTraversal(node.right, list);
        }
    }
}

/// Source : https://leetcode.com/problems/binary-tree-inorder-traversal/
/// Author : liuyubobobo
/// Time   : 2017-11-17

import java.util.ArrayList;
import java.util.List;
import java.util.Stack;

// My Non-Recursive
// Time Complexity: O(n), n is the node number in the tree
// Space Complexity: O(h), h is the height of the tree
public class Solution2 {

    private class Command{
        String s;   // go, print
        TreeNode node;
        Command(String s, TreeNode node){
            this.s = s;
            this.node = node;
        }
    };

    public List<Integer> inorderTraversal(TreeNode root) {

        ArrayList<Integer> res = new ArrayList<Integer>();
        if(root == null)
            return res;

        Stack<Command> stack = new Stack<Command>();
        stack.push(new Command("go", root));
        while(!stack.empty()){
            Command command = stack.pop();

            if(command.s.equals("print"))
                res.add(command.node.val);
            else{
                assert command.s.equals("go");
                if(command.node.right != null)
                    stack.push(new Command("go",command.node.right));
                stack.push(new Command("print", command.node));
                if(command.node.left != null)
                    stack.push(new Command("go",command.node.left));
            }
        }
        return res;
    }

}

/// Source : https://leetcode.com/problems/binary-tree-inorder-traversal/
/// Author : liuyubobobo
/// Time   : 2018-05-30

import java.util.ArrayList;
import java.util.List;
import java.util.Stack;

// Classic Non-Recursive algorithm for inorder traversal
// Time Complexity: O(n), n is the node number in the tree
// Space Complexity: O(h), h is the height of the tree
public class Solution3 {

    public List<Integer> inorderTraversal(TreeNode root) {

        ArrayList<Integer> res = new ArrayList<Integer>();
        if(root == null)
            return res;

        Stack<TreeNode> stack = new Stack<>();
        TreeNode cur = root;
        while(cur != null || !stack.empty()){

            while(cur != null){
                stack.push(cur);
                cur = cur.left;
            }

            cur = stack.pop();
            res.add(cur.val);
            cur = cur.right;
        }
        return res;
    }
}

/// Source : https://leetcode.com/problems/binary-tree-inorder-traversal/
/// Author : liuyubobobo
/// Time   : 2018-05-30

import java.util.ArrayList;
import java.util.List;
import java.util.Stack;

// Another Classic Non-Recursive algorithm for inorder traversal
// Time Complexity: O(n), n is the node number in the tree
// Space Complexity: O(h), h is the height of the tree
public class Solution4 {

    public List<Integer> inorderTraversal(TreeNode root) {

        ArrayList<Integer> res = new ArrayList<Integer>();
        if(root == null)
            return res;

        Stack<TreeNode> stack = new Stack<>();
        TreeNode cur = root;
        while(cur != null || !stack.empty()){

            if(cur != null){
                stack.push(cur);
                cur = cur.left;
            }
            else{
                cur = stack.pop();
                res.add(cur.val);
                cur = cur.right;
            }
        }
        return res;
    }
}

/// Source : https://leetcode.com/problems/binary-tree-inorder-traversal/
/// Author : liuyubobobo
/// Time   : 2018-05-30

import java.util.ArrayList;
import java.util.List;
import java.util.Stack;

// Inorder Morris Traversal
// Time Complexity: O(n), n is the node number in the tree
// Space Complexity: O(1)
public class Solution5 {

    public List<Integer> inorderTraversal(TreeNode root) {

        ArrayList<Integer> res = new ArrayList<Integer>();
        if(root == null)
            return res;

        TreeNode cur = root;
        while(cur != null){

            if(cur.left == null){
                res.add(cur.val);
                cur = cur.right;
            }
            else{
                TreeNode prev = cur.left;
                while(prev.right != null && prev.right != cur)
                    prev = prev.right;

                if(prev.right == null){
                    prev.right = cur;
                    cur = cur.left;
                }
                else{
                    prev.right = null;
                    res.add(cur.val);
                    cur = cur.right;
                }
            }
        }
        return res;
    }
}