标签：path sum any obj problem 左右 相加 最大值 答案 bsp
求二叉树的最大路径和
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.
For example:
Given the below binary tree,1 / 2 3
Return
6.
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
# 该题采用递归求解，不过递归时应注意局部的最大路径和返回值不相等，返回值只能是单一路径，而最大路径可以使左右相加
class Solution(object):
def maxPathSum(self, root):
"""
:type root: TreeNode
:rtype: int
"""
self.maxPath = -sys.maxint
self.helper(root)
return self.maxPath
def helper(self, node):
if node == None:
return 0
lMax = self.helper(node.left)
rMax = self.helper(node.right)
# 局部最大路径 = 根节点值 + 左、右最大路径（必须为正才能加）
MaxSum = node.val
if lMax > 0:
MaxSum += lMax
if rMax > 0:
MaxSum += rMax
self.maxPath = max(self.maxPath, MaxSum)
# 返回值为单一路径最大值，如果左、右最大路径均为负，则返回根节点值
return max(lMax, rMax, 0) + node.val
LeetCode 124. Binary Tree Maximum Path Sum
标签：path sum any obj problem 左右 相加 最大值 答案 bsp
原文地址：http://www.cnblogs.com/LiCheng-/p/6892562.html
