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
这个题比之前的要难一些,现在的路径可以是中间的其中的一段都是可以的。
# 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()
上面的解法我是想找到的同时把路径也保存下来
如果不保存结果的话
# 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





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