# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
def postorderTraversal(self, root):
"""
:type root: TreeNode
:rtype: List[int]
"""
if root == None:
return []
stack = []
res = []
to_apd = root
lastVisit = None
while to_apd != None or len(stack) != 0:
if to_apd != None:
# All the appending work is done here.
stack.append(to_apd)
to_apd = to_apd.left
else:
n = stack[-1]
if n.right != None and n.right != lastVisit:
# If n has a right node and it is not visited, append the right node.
to_apd = n.right
else:
# Else, consume n
del stack[-1]
res.append(n.val)
lastVisit = n
return res
# You can simply reverse the preoder of the tree to get the post oder.
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
def postorderTraversal(self, root):
"""
:type root: TreeNode
:rtype: List[int]
"""
if root == None:
return []
stack = [root]
res = []
while len(stack) != 0:
curNode = stack[-1]
del stack[-1]
res.append(curNode.val)
if curNode.left:
stack.append(curNode.left)
if curNode.right:
stack.append(curNode.right)
return res[::-1]
https://leetcode.com/problems/binary-tree-postorder-traversal/#/description
Given a binary tree, return the postorder traversal of its nodes' values.
For example:
Given binary tree {1,#,2,3}
,
1
\
2
/
3
return [3,2,1]
.
Note: Recursive solution is trivial, could you do it iteratively?