144. Binary Tree Preorder Traversal

Difficulty: Easy

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

1. Recursive method

$O(n)$ runtime,

$O(n)$ space

python

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def preorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
        result = []
        def preorder(node):
            if node is not None:
                result.append(node.val)
                preorder(node.left)
                preorder(node.right)
        
        preorder(root)
        return result

2. Iterative method with Stack

$O(n)$ runtime,

$O(n)$ space

python

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def preorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
        stack, result = [], []
        node = root
        while True:
            while node is not None:
                result.append(node.val)
                stack.append(node)
                node = node.left
            if not stack:
                return result
            node = stack.pop()
            node = node.right

python

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def preorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
        stack, result = [root], []
        while stack:
            node = stack.pop()
            if node is not None:
                result.append(node.val)
                stack.append(node.right)
                stack.append(node.left)
        
        return result

3. Morris Traversal

$O(n)$ runtime,

$O(1)$ space

python

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def preorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
        result = []
        node = root
        while node is not None:
            if node.left is None:
                result.append(node.val)
                node = node.right
            else:
                # Find the inorder predecessor of current node
                pred = node.left
                while pred.right is not None and pred.right is not node:
                    pred = pred.right
                
                # If the right child of inorder predecessor already points to
                # the current node, restore the original tree and update the 
                # current node to it's right child
                if pred.right is node:
                    pred.right = None
                    node = node.right
                
                # Make current node as right child of its inorder predecessor
                else:
                    result.append(node.val)
                    pred.right = node
                    node = node.left
        
        return result

Ref: https://www.796t.com/content/1545897069.html

tags: leetcode