LeetCode 124. Binary Tree Maximum Path Sum

原题

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.

解题思路

1. 递归求解
2. 简化问题，先考虑全是正数的情况下的解法
3. 重点：需要两个最大值，一个是当前树的最大路径，即当前节点值加上左右子树的节点值；另一个是当前树向父节点提供的最大路径，这个值应该是根节点的值加上路径最长的子树那边的最大路径，向上层递归函数返回这个值。
4. 于是考虑到负值的话即为：需要两个最大值，一个是当前树的最大路径，即当前节点值加上左右子树的路径和（如果左右子树路径和为正的话）；另一个是当前树向父节点提供的最大路径，这个值应该是根节点的值加上路径最长的子树那边的最大路径，向上层递归函数返回这个值。

完整代码

# 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
        """
        if root == None:
            return 0
        self.maxSum = -sys.maxint
        self._maxPathSum(root)
        return self.maxSum
        
    def _maxPathSum(self, node):
        """
        :type node: TreeNode
        :rtype: 对于上层根的最大路径和
        """
        if node == None:
            return 0
        node_max = node.val
        lPathSum, rPathSum = self._maxPathSum(node.left), self._maxPathSum(node.right)
        if lPathSum > 0:
            node_max += lPathSum
        if rPathSum > 0:
            node_max += rPathSum
        # 求此节点当为根节点时的最大路径和
        if self.maxSum < node_max:
            self.maxSum = node_max
        # 返回对于上层根节点的最大路径和    
        return max(node.val, node.val + lPathSum, node.val + rPathSum)