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]
]
Solutions
1. DFS-递归
思路是一样的，
# Time Complexity: O(n)
# Sapce Complexity: O(1)
# 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) -> List[List[int]]:
if not root:
return []
res = []
self.dfs(root, sum, [], res)
return res
def dfs(self, root, sum, path, res):
if not root.left and not root.right and sum == root.val:
path.append(root.val)
res.append(path)
cur_sum = sum - root.val
if root.left:
self.dfs(root.left, cur_sum, path + [root.val], res)
if root.right:
self.dfs(root.right, cur_sum, path + [root.val], res)
# Runtime: 44 ms, faster than 99.04% of Python3 online submissions for Path Sum II.
# Memory Usage: 19.1 MB, less than 6.90% of Python3 online submissions for Path Sum II.
2. 迭代
# Time Complexity: O(n)
# Sapce Complexity: O(1)
# 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) -> List[List[int]]:
if not root:
return []
if not root.left and not root.right and sum == root.val:
return [[root.val]]
cur_sum = sum - root.val
tmp = self.pathSum(root.left, cur_sum) + self.pathSum(root.right, cur_sum)
return [[root.val] + path for path in tmp]
# Runtime: 56 ms, faster than 61.99% of Python3 online submissions for Path Sum II.
# Memory Usage: 15.6 MB, less than 37.93% of Python3 online submissions for Path Sum II.