144. Binary Tree Preorder Traversal

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https://leetcode.com/problems/binary-tree-preorder-traversal/

Time complexity: O(N), Space complexity: O(N)

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
// Recursion
class Solution_1 {
    List<Integer> ls=new ArrayList<>();
    public List<Integer> preorderTraversal(TreeNode root) {
        if(root!=null){
            ls.add(root.val);
            preorderTraversal(root.left);
            preorderTraversal(root.right);
        }
        return ls;
    }
}

// Iteration
class Solution_2 {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> ls=new ArrayList<>();
        LinkedList<TreeNode> stack=new LinkedList<>();
        stack.push(root);
        while(!stack.isEmpty()){
            TreeNode node=stack.pop(); 
            if(node!=null){
                ls.add(node.val);
                // be careful about the order. 
                // Firstly push the right branch then push the left.
                stack.push(node.right);
                stack.push(node.left);
            }
        }
        return ls;
    }
}

Morris Traversal [Time complexity: O(N), Space complexity: O(1)]
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class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> ls=new ArrayList<>();
        while(root!=null){
            if(root.left==null){
                ls.add(root.val);
                root=root.right;
            }else{
                TreeNode predecessorNode=root.left;
                while(predecessorNode.right!=null&&predecessorNode.right!=root) 
                    predecessorNode=predecessorNode.right;
                if(predecessorNode.right==null){
                    ls.add(root.val);
                    predecessorNode.right=root;
                    root=root.left;
                }else{
                    predecessorNode.right=null;
                    root=root.right;
                }
            } 
        }
        return ls;
    }
}

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