leetcode_binary tree postorder traversal

Binary Tree Postorder Traversal

Given a binary tree, return the postorder traversal of its nodes’ values.
(后序遍历二叉树)

Follow up: Recursive solution is trivial, could you do it iteratively?

Example:

1. 递归

具体实现方法如下:

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# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None

class (object):
def helper(self, root, res):
if not root:
return res

self.helper(root.left, res)
self.helper(root.right, res)
res.append(root.val)

def postorderTraversal(self, root):
"""
:type root: TreeNode
:rtype: List[int]
"""
if not root:
return []

result = []
self.helper(root, result)

return result

2. 迭代 & 栈

由于先序遍历的顺序是根-左-右,后序遍历的顺序是左-右-根,因此可以修改先序遍历的遍历顺序为根-右-左,然后将最终结果倒序即可。具体实现方法如下:

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# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None

class (object):
def postorderTraversal(self, root):
"""
:type root: TreeNode
:rtype: List[int]
"""
if not root:
return []
result, stack = [], []
p = root

while stack or p:
if p:
result.append(p.val)
stack.append(p.left)
p = p.right
else:
p = stack.pop()

return result[::-1]