【DFS】leetcode每日一题—437.路径总和III

题目：
给定一个二叉树的根节点 root ，和一个整数 targetSum ，求该二叉树里节点值之和等于 targetSum 的 路径 的数目。

路径 不需要从根节点开始，也不需要在叶子节点结束，但是路径方向必须是向下的（只能从父节点到子节点）。

思路：
参考官方题解

解答：
方法一：

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right

class Solution:
    def pathSum(self, root: TreeNode, targetSum: int) -> int:
        def rootSum(root, targetSum):
            if root is None:
                return 0

            ret = 0
            if root.val == targetSum:
                ret += 1

            ret += rootSum(root.left, targetSum - root.val)
            ret += rootSum(root.right, targetSum - root.val)
            return ret
        
        if root is None:
            return 0
            
        ret = rootSum(root, targetSum)
        ret += self.pathSum(root.left, targetSum)
        ret += self.pathSum(root.right, targetSum)
        return ret

方法二：利用前缀和，减小计算量
prefix={前缀和：前缀和出现的次数}

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def pathSum(self, root: TreeNode, targetSum: int) -> int:
        prefix = collections.defaultdict(int)
        prefix[0] = 1

        def dfs(root, curr):
            if not root:
                return 0
            
            ret = 0
            curr += root.val
            ret += prefix[curr - targetSum]
            prefix[curr] += 1
            #print(prefix)
            ret += dfs(root.left, curr)
            ret += dfs(root.right, curr)
            prefix[curr] -= 1
            #print(prefix)

            return ret

        return dfs(root, 0)