leetcode 437 路径总和III

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）。

示例 1：

输入：root = [10,5,-3,3,2,null,11,3,-2,null,1], targetSum = 8
输出：3
解释：和等于 8 的路径有 3 条，如图所示。

示例 2：

输入：root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22
输出：3

 自己写的...没通过

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution(object):
    def pathSum(self, root, targetSum):
        """
        :type root: TreeNode
        :type targetSum: int
        :rtype: int
        """
        def singlePathSum(pathValList,count, targetSum, answerList):
            i = 1 # 步长
            while i <= len(pathValList):
                j = 0
                while j + i <= len(pathValList):
                    k = j
                    sum = 0
                    tmpArr = []
                    while k < j + i:
                        sum += pathValList[k]
                        tmpArr.append(pathValList[k])
                        k += 1
                    if sum == targetSum:
                        count += 1
                        print(tmpArr)
                        answerList.append(tmpArr)
                    j += 1
                i += 1
            return count
        def goNextlevel(root, pathValList, count, targetSum, answerList):
            print("curVal=", root.val)
            pathValList.append(root.val)
            if not root.left and not root.right:
                count = singlePathSum(pathValList,count, targetSum, answerList)
                return count
            if root.left:
                count = goNextlevel(root.left, pathValList,  count, targetSum, answerList)
                pathValList.pop()
            if root.right:
                count = goNextlevel(root.right, pathValList, count, targetSum, answerList)
                pathValList.pop()
            return count
        count = 0
        if not root:
            return count
        else:
            if not root.left and not root.right:
                if root.val == targetSum:
                    count += 1
                return count
        pathValList = []
        answerList = []
        pathValList.append(root.val)
        if root.left:
            count = goNextlevel(root.left, pathValList, count, targetSum, answerList)
        if root.right:
            count = goNextlevel(root.right, pathValList, count, targetSum, answerList)
        answerList = [x for i, x in enumerate(answerList) if x not in answerList[:i]]
        return len(answerList)

题解：每向下一层就把targetSum的值减掉当前节点的值

其实就是把每个节点作为根节点，分治法再计算每个子树的符合条件的路径个数，求和

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