Given the root of a binary tree, return the postorder traversal of its nodes' values.
Example 1:
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Input: root = [1,null,2,3]
Output: [3,2,1]
Example 2:
Input: root = []
Output: []
Example 3:
Input: root = [1]
Output: [1]
Example 4:
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Input: root = [1,2]
Output: [2,1]
Example 5:
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Input: root = [1,null,2]
Output: [2,1]
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 postorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
def postorder(node: Optional[TreeNode], lst: List[int]):
if not node:
return
postorder(node.left, lst)
postorder(node.right, lst)
lst.append(node.val)
result = []
postorder(root, result)
return result
''' Iterative '''
class Solution:
def postorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
if root is None:
return []
stack, res = [root], []
while stack:
node = stack.pop()
if node:
# pre-order, right first
res.append(node.val)
stack.append(node.left)
stack.append(node.right)
# reverse result
return res[::-1]
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