Given a non-empty 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.
Example 1:
Input: [1,2,3] 1 / \ 2 3 Output: 6
Example 2:
Input: [-10,9,20,null,null,15,7] -10 / \ 9 20 / \ 15 7 Output: 42
LeetCode:链接
maxSum用于记录最大路径和,当level为0时返回它。maxPathSum(root)函数返回的是经过root的路径最大和, 可以只是root, 也可以是左子树某一部分+root或者root+右子树某一部分。这个函数是可以返回0的, 也就是说currRoot, currRoot+左子树, currRoot+右子树的值都小于等于0, 对max path sum没有贡献, 之后就不再考虑了。
# 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 maxPathSum(self, root):
"""
:type root: TreeNode
:rtype: int
"""
self.maxSum = float("-inf")
self.dfs(root)
return self.maxSum
def dfs(self, root):
if not root:
return 0
vleft = self.dfs(root.left)
vright = self.dfs(root.right)
self.maxSum = max(self.maxSum, vleft + vright + root.val)
return max(root.val + vleft, root.val + vright, 0)