144. Binary Tree Preorder Traversal (DFS +Stack) 20.3.7 Medium

Given a binary tree, return the preorder traversal of its nodes' values.

Example:

Input: [1,null,2,3]
   1
    \
     2
    /
   3

Output: [1,2,3]
Follow up: Recursive solution is trivial, could you do it iteratively?

Solution: ----------------------------------------------------------- DFS 
----------------------------------------------------------- O(n) T, O(n) S
class Solution(object):
    def preorderTraversal(self, root):
        res = []
        self.traver_help(res, root)
        return res
        """
        :type root: TreeNode
        :rtype: List[int]
        """
    def traver_help(self, res, root):
         if not root is None:
            res.append(root.val)
            self.traver_help(res, root.left)
            self.traver_help(res, root.right)
'''
Tomorrow is midterm,
So just refresh the resursion solution for DFS traversal,
the iterative solution will be followed up later on.
'''

Solution: ----------------------------------------------------------- Stack
----------------------------------------------------------- O(n) T, O(n) S
# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def preorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        if not root:
            return None
        stack, res = [root], []
        while stack:
            pop_element = stack.pop()
            res.append(pop_element.val)              # Append the node.val first -> preorder
                                                     # Make use of the First in Last out property for stack,
                                                     # So, we append node.right before append node.left
                                                     # In the next interation, node = node.left and continue
            if pop_element.right:                    # IMPORTANT !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
                                                     # In both stack and queue question, everytime we want to append, 
                                                     # we should test whether it is None, because None is meaningless and
                                                     # may bring "NoneType object has no attribute" error in next iteration
                stack.append(pop_element.right)
            if pop_element.left:
                stack.append(pop_element.left)
        return res


Java Version:
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        if (root == null) {
            return res;
        }
        dfs(root, res);
        return res;
    }
    
    public void dfs(TreeNode root, List<Integer> res) {
        if (root == null) {
            return;
        }
        res.add(root.val);
        dfs(root.left, res);
        dfs(root.right, res);
    }
}


Java Version for solution no.2:
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        if (root == null) {
            return res;
        }
        Stack<TreeNode> s = new Stack<>();
        s.add(root);
        while (s.size() != 0) {
            TreeNode popNode = s.pop();
            res.add(popNode.val);
            if (popNode.right != null) {
                s.add(popNode.right);
            }
            if (popNode.left != null) {
                s.add(popNode.left);
            }
        }
        return res;
    }
}