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原题
https://leetcode.com/problems/path-sum-ii/
解法1
DFS. Base case是当root为空时, 返回空列表. 定义dfs函数, base case是root为空时, 直接返回. 由于题意要求从根节点到叶子节点, 那么我们判断当root为叶子节点时, 将值加入path, 然后将path加入res.
Time: O(n)
Space: O(1)
代码
# 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]]':
def dfs(root, sum, path, res):
if not root:
return
if root.left is None and root.right is None and root.val == sum:
path.append(root.val)
res.append(path)
dfs(root.left, sum - root.val, path + [root.val], res)
dfs(root.right, sum - root.val, path + [root.val], res)
res = []
if not root: return res
dfs(root, sum, [], res)
return res
解法2
BFS
代码
# 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]]':
res = []
if not root: return res
q = [(root, root.val, [root.val])]
while q:
node, val, path = q.pop(0)
if node.left is None and node.right is None and val == sum:
res.append(path)
if node.left:
q.append((node.left, val+node.left.val, path+[node.left.val]))
if node.right:
q.append((node.right, val+node.right.val, path+[node.right.val]))
return res