leetcode437 -- Path Sum III

# 小郑之家~

### 题目



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 pathSum(self, root: TreeNode, sum: int) -> int:
if not root:
return 0
res = []
self.helper(root, sum , 0, res, [])
print(res)
return len(res)

def helper(self, root, target, curSum, res, out):
if not root:
return

curSum += root.val
out.append(root.val)

if curSum == target:
res.append(out[:])

temp = curSum
for i in range(len(out)-1):  ## 这是要看去掉一些能不能行，要保留一个。
temp -= out[i]
Temp = out[:]
del(Temp[i])
if temp == target:
res.append(Temp)
self.helper(root.left, target, curSum, res,out[:])
self.helper(root.right, target, curSum, res, out[:])
out.pop()



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

class Solution:
def pathSum(self, root: TreeNode, sum: int) -> int:
if not root:
return 0

return self.helper(root, sum , 0, 0, [])

def helper(self, root, target, curSum, res, out):
if not root:
return res  # 注意这里是返回res.

curSum += root.val
out.append(root.val)

if curSum == target:
res += 1

temp = curSum
for i in range(len(out)-1):
temp -= out[i]
Temp = out[:]
del(Temp[i])
if temp == target:
res += 1

res = self.helper(root.left, target, curSum, res,out[:])
res = self.helper(root.right, target, curSum, res, out[:])
out.pop()
return res