题目描述：

Given a binary tree, find the maximum path sum.

For this problem, a path is defined as any sequence of nodes from some starting node to any node in the tree along the parent-child

connections. The path must contain at least one node and does not need
to go through the root.

举例：

Given the below binary tree,

   1
  / 
 2   3
Return 6.

题目分析： 找从任意节点出发的任意路径的最大长度。  每个node都有可能是其他路径上的node，这种情况要max(left，right)。如此循环。  每个node都有可能作为中心node，此时要max(左侧之前的路径最长长度，左侧之前的路径最长长度，此node为中心时候的长度)
将这个分析单元递归封装，即可实现目标。
# Definition for a binary tree node.
class TreeNode:
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None

class Solution:
    def dfs(self,node):
        ls = rs = None
        lmv = rmv = 0
        if node.left:
            lmv,ls=self.dfs(node.left)
            lmv=max(lmv,0)
        if node.right:
            rmv,rs=self.dfs(node.right)
            rmv=max(rmv,0)
        # print(lmv,rmv,ls,rs)
        mv=node.val+max(lmv,rmv)
        sv=node.val+lmv+rmv
        # mv=node.val
        trans_list=[elem for elem in [sv,ls,rs] if elem]
        if not trans_list:
            trans_list=[0]
        return mv,max(trans_list)

    def maxPathSum(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        if not root:
            return
        mv,smv=self.dfs(root)
        return max(mv,smv)

if __name__=='__main__':
    tn=TreeNode(2)
    tn1=TreeNode(-1)
    tn2=TreeNode(-2)
    tn.left=tn1
    tn.right=tn2