Python算法练习 10.16
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