Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path 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 1
return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.
思路
这道题我们使用递归进行解决，递归结束条件为节点为空时返回False，还有一个就是就是当前节点为叶子节点(叶子节点就是左右节点为空)，并且sum是否为0时返回True。
解决代码
1 # Definition for a binary tree node. 2 # class TreeNode(object): 3 # def __init__(self, x): 4 # self.val = x 5 # self.left = None 6 # self.right = None 7 8 class Solution(object): 9 def hasPathSum(self, root, sum): 10 """ 11 :type root: TreeNode 12 :type sum: int 13 :rtype: bool 14 """ 15 if not root: # 空节点直接返回 16 return False 17 if root and not root.left and not root.right and root.val == sum: # 当前节点为叶子节点，并且sum为0 18 return True 19 return self.hasPathSum(root.left, sum-root.val) or self.hasPathSum(root.right, sum-root.val) # 左右节分别进行判断
如果我们需要得到满足路径的值呢？
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]
]
思路
和上面那道题的解决方式一样，只不过多一个path和res来记录节点路径和结果列表。还是使用递归。
解决代码
1 # Definition for a binary tree node.
2 # class TreeNode(object):
3 # def __init__(self, x):
4 # self.val = x
5 # self.left = None
6 # self.right = None
7
8 class Solution(object):
9 def pathSum(self, root, sum):
10 """
11 :type root: TreeNode
12 :type sum: int
13 :rtype: List[List[int]]
14 """
15 if not root:
16 return []
17 res = []
18 self.path(root, sum, res, [])
19 return res
20
21 def path(self, root, sum, res, path):
22 if not root: # 为空直接返回
23 return
24 if not root.left and not root.right and root.val == sum: # 判断该叶子节点的路径是否满足条件
25 path.append(root.val)
26 res.append(path)
27 return
28 self.path(root.left, sum-root.val, res, path+[root.val]) # 该节点的左子树
29 self.path(root.right, sum-root.val, res,path+[root.val]) # 该节点的右子树