Given the root of a binary tree, return the preorder traversal of its nodes' values.
Example 1:
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Input: root = [1,null,2,3]
Output: [1,2,3]
Example 2:
Input: root = []
Output: []
Example 3:
Input: root = [1]
Output: [1]
Example 4:
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Input: root = [1,2]
Output: [1,2]
Example 5:
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Input: root = [1,null,2]
Output: [1,2]
Constraints:
- The number of nodes in the tree is in the range
[0, 100].
- -100 <=
Node.val <= 100
Follow up: Recursive solution is trivial, could you do it iteratively?
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# 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
''' Recursive '''
class Solution:
def preorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
def preorder(node: Optional[TreeNode], lst: List[int]):
if not node:
return
lst.append(node.val)
preorder(node.left, lst)
preorder(node.right, lst)
result = []
preorder(root, result)
return result
''' Iterative '''
class Solution:
def preorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
if root is None:
return []
stack, output = [root, ], []
while stack:
root = stack.pop()
if root:
output.append(root.val) # Add before going to children
if root.right:
stack.append(root.right)
if root.left:
stack.append(root.left)
return output
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