卖萌的弱渣

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Binary Tree Maximum Path Sum

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 does not need to go through the root.

Example:

Given the below binary tree,

1
2
3

   1
  / \
 2   3

Return 6.

Solution

  • DFS，注意节点可能是负值
(Binary-Tree-Maximum-Path-Sum.py) download
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# 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.ans = 0
        def dfs(node):
            if node == None:
                return 0
            leftMax = dfs(node.left)
            rightMax = dfs(node.right)
            # 比较leftmax+rightmax+node.val 和self.ans (因为node.val可能是负数)
            self.ans = max(max(leftMax,0)+max(rightMax,0)+node.val, self.ans)
            # 因为leftMax,rightMax可能是负数,我们只要root.val
            return max(leftMax，rightMax,0)+node.val
        return self.ans