Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.
Note: A leaf is a node with no children.
Example:
Given the below binary tree and sum = 22,
5
/ \
4 8
/ / \
11 13 4
/ \ / \
7 2 5 1
Return:
[
[5,4,11,2],
[5,8,4,5]
]
class Solution:
def pathSum(self, root: TreeNode, sum: int) -> List[List[int]]:
ans = []
self.dfs(root, sum, [], ans)
return ans
def dfs(self, root, sum, path, ans):
if not root:
return
path.append(root.val)
if root.val == sum and not root.right and not root.left:
ans.append(list(path))
self.dfs(root.left, sum - root.val, path, ans)
self.dfs(root.right, sum - root.val, path, ans)
path.pop()