leetcode(404)：Sum of Left Leaves

题目：Find the sum of all left leaves in a given binary tree.

example:

3

/ \

9 20

/ \

15 7

There are two left leaves in the binary tree, with values 9 and 15 respectively. Return 24.

题目分析：求所有左叶子节点的和，然后判断是否为左叶子节点，在这个地方容易出现的问题：第一，在于单个节点的情况的漏判；第二是对于左叶子节点的判断，不能按照叶子节点的判断来走，必须按照左叶子节点的判断来处理；第三是对于如何计数和的处理，这个也需要仔细的思考。

python代码实现:

class Solution(object):
def sumOfLeftLeaves(self, root):
"""
:type root: TreeNode
:rtype: int
"""
if root == None:
return 0
if root.left == None and root.right == None:
return 0
root.val = 0
if root.left != None:
if root.left.left == None and root.left.right == None:
root.val += root.left.val
else:
root.val += self.sumOfLeftLeaves(root.left)
root.val += self.sumOfLeftLeaves(root.right)
return root.val

重新思考问题，重新复现这个问题的求解，代码如下：

class Solution(object):
def sumOfLeftLeaves(self, root):
"""
:type root: TreeNode
:rtype: int
"""
if not root or (not root.left and not root.right):
return 0
if root.left:
if not root.left.left and not root.left.right:
return root.left.val + self.sumOfLeftLeaves(root.right)
return self.sumOfLeftLeaves(root.left) + self.sumOfLeftLeaves(root.right)

大佬的实现代码：

class Solution(object):
def sumOfLeftLeaves(self, root):
if not root: return 0
if root.left and not root.left.left and not root.left.right:
return root.left.val + self.sumOfLeftLeaves(root.right)
return self.sumOfLeftLeaves(root.left) + self.sumOfLeftLeaves(root.right)   # isn't leave

好吧，很开心，跟大佬的想法一样，只是还是老问题，就是大佬们的代码相对而言，更加简洁，对于多个条件的判断，然后把他们放在一起，更加清楚，方便。