Binary Tree Postorder Traversal*
Given a binary tree, return the postorder traversal of its nodes' values.
Example
Given binary tree {1,#,2,3},
1
\
2
/
3
return [3,2,1]
Solution
(1) 首选是搜索向下,如果有左,就不停的把左节点压入占栈,如果左空,则判断右,右不空,把右压入堆栈
(2) 搜索向上,有两种情况, 一种是左节点回到父节点,一种是右节点回到父节点,由于是,左右根,所以如果从左节点回到根,需要由跟节点转移到右节点。
(3) 如果节点在左右节点prev == curr, 或者 curr.right = prev, 右节点返回跟节点,这是就可以按顺序压入列表,并且弹出堆栈。
"""
Definition of TreeNode:
class TreeNode:
def __init__(self, val):
self.val = val
self.left, self.right = None, None
"""
class Solution:
"""
@param root: The root of binary tree.
@return: Postorder in ArrayList which contains node values.
"""
def postorderTraversal(self, root):
# write your code here
if root is None:
return []
stack = [root]
postorderResult = []
prev = None
curr = None
while stack:
curr = stack[-1]
if prev is None or prev.left == curr or prev.right == curr: # traverse down the tree
if curr.left:
stack.append(curr.left)
elif curr.right: #非常重要,如果左子树为空的时候,这时候才能考虑是从右面过来的。
stack.append(curr.right)
elif curr.left == prev: # traverse up from the left of the tree then need to go to right
if curr.right:
stack.append(curr.right)
else: # the end of node will be added when prev == curr , that means the end.
postorderResult.append(curr.val)
stack.pop()
prev = curr
return postorderResult
Last updated
Was this helpful?