递归与二叉树_leetcode437

# 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 pathSum(self, root, sum):
        """
        :type root: TreeNode
        :type sum: int
        :rtype: int
        """
        if not root :
            return 0

        return self.findPath(root,sum) + self.pathSum(root.left,sum) + self.pathSum(root.right,sum)

    def findPath(self,root,sum):

        if not root :
            return 0

        res = 0

        if root.val == sum :
            res += 1

        res += self.findPath(root.left,sum-root.val)
        res += self.findPath(root.right,sum-root.val)

        return res

递归与二叉树_leetcode437

原文：https://www.cnblogs.com/lux-ace/p/10546817.html