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0145-Binary-Tree-Postorder-Traversal.py
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0145-Binary-Tree-Postorder-Traversal.py
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'''
Given the root of a binary tree, return the postorder traversal of its nodes' values.
Example 1:
Input: root = [1,null,2,3]
Output: [3,2,1]
Example 2:
Input: root = []
Output: []
Example 3:
Input: root = [1]
Output: [1]
Example 4:
Input: root = [1,2]
Output: [2,1]
Example 5:
Input: root = [1,null,2]
Output: [2,1]
Constraints:
The number of the 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?
'''
# 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: TreeNode) -> List[int]:
self.res = []
self.recursive(root)
return self.res
def recursive(self, root):
if not root:
return
self.recursive(root.left)
self.recursive(root.right)
self.res.append(root.val)
# Iterative 1
class Solution:
def postorderTraversal(self, root: TreeNode) -> List[int]:
res = []
stack = [(root, False)]
while stack:
node, visited = stack.pop()
if node and visited:
res.append(node.val)
elif node:
stack.append((node, True))
stack.append((node.right, False))
stack.append((node.left, False))
return res
# Iterative 2
class Solution:
def postorderTraversal(self, root: TreeNode) -> List[int]:
res = []
stack = []
cur = root
while stack or cur:
if cur:
stack.append(cur)
cur = cur.left
else:
node = stack[-1].right
if not node:
node = stack.pop()
res.append(node.val)
while stack and stack[-1].right == node:
node = stack.pop()
res.append(node.val)
else:
cur = node
return res