# # 【LeetCode】199. Binary Tree Right Side View 解题报告（Python）

## # 题目描述：

Given a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom.

``````For example:
Given the following binary tree,

1            <---
/   \
2     3         <---
\     \
5     4       <---

You should return [1, 3, 4].
``````

## # 解题方法

``````# 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 rightSideView(self, root):
"""
:type root: TreeNode
:rtype: List[int]
"""
res = []
self.levelOrder(root, 0, res)
return [level[-1] for level in res]

def levelOrder(self, root, level, res):
if not root: return
if len(res) == level: res.append([])
res[level].append(root.val)
if root.left: self.levelOrder(root.left, level + 1, res)
if root.right: self.levelOrder(root.right, level + 1, res)
``````

``````# 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 rightSideView(self, root):
"""
:type root: TreeNode
:rtype: List[int]
"""
res = []
if not root: return res
queue = collections.deque()
queue.append(root)
while queue:
res.append(queue[-1].val)
for i in range(len(queue)):
node = queue.popleft()
if node.left:
queue.append(node.left)
if node.right:
queue.append(node.right)
return res
``````

## # 日期

2018 年 3 月 14 日 --霍金去世日