437. Path Sum III

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本文介绍了一种算法，用于查找给定二叉树中所有路径之和等于特定值的数量。路径可以从任意节点开始，但只能向下遍历。文章提供了两种解决方案，并详细解释了递归深度优先搜索的方法。

题目描述：

You are given a binary tree in which each node contains an integer value.

Find the number of paths that sum to a given value.

The path does not need to start or end at the root or a leaf, but it must go downwards (traveling only from parent nodes to child nodes).

The tree has no more than 1,000 nodes and the values are in the range -1,000,000 to 1,000,000.

Example:

root = [10,5,-3,3,2,null,11,3,-2,null,1], sum = 8

      10
     /  \
    5   -3
   / \    \
  3   2   11
 / \   \
3  -2   1

Return 3. The paths that sum to 8 are:

1.  5 -> 3
2.  5 -> 2 -> 1
3. -3 -> 11

# Definition for a binary tree node.

# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def path(self, root, sum):
        """
        :type root: TreeNode
        :type sum: int
        :rtype: int
        """
        self.num = 0
        if root:
            if root.val == sum:
                self.num+=1
            self.num+=self.path(root.left,sum-root.val)
            self.num+=self.path(root.right,sum-root.val)
        return self.num
    def pathSum(self,root,sum):
        if root is None:
            return 0
        return self.path(root,sum)+self.pathSum(root.left,sum)+self.pathSum(root.right,sum)

思路：自己做的太麻烦了，修改了好多次，不过思想还是自己容易理解的，就是依次从根节点开始深度遍历，知道到达子节点，看看途中一共有多少的路径满足要求。进入过许多误区，主要是对题目的不理解，题目中可能出现负数的情况，另外，自己提交的答案并不好，说明优化的地方还有好多。运行时间太长

class Solution:
    def pathSum(self, root, target):
        # define global result and path
        self.result = 0
        cache = {0:1}
        
        # recursive to get result
        self.dfs(root, target, 0, cache)
        
        # return result
        return self.result
    
    def dfs(self, root, target, currPathSum, cache):
        # exit condition
        if root is None:
            return  
        # calculate currPathSum and required oldPathSum
        currPathSum += root.val
        oldPathSum = currPathSum - target
        # update result and cache
        self.result += cache.get(oldPathSum, 0)
        cache[currPathSum] = cache.get(currPathSum, 0) + 1
        
        # dfs breakdown
        self.dfs(root.left, target, currPathSum, cache)
        self.dfs(root.right, target, currPathSum, cache)
        # when move to a different branch, the currPathSum is no longer available, hence remove one. 
        cache[currPathSum] -= 1