112 path sum

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.

sum = 22

1
2
3
4
5
6
7
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

这一题也是利用深度优先遍历(DFS)的递归方法来进行查找。

需要注意的是,只有自根节点到叶子节点的路径才算是一条符合题意的路径,其它情况都不需要额外考虑了。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class (object):
def hasPathSum(self, root, sum):
"""
:type root: TreeNode
:type sum: int
:rtype: bool
"""
if (root):
return self.DFS(root, sum)
else:
return False
def DFS(self, root, sum):
if (root):
val = root.val
sum -= val
if(root.left is None and root.right is None):
if (sum == 0):
return True
else:
return False
else:
return self.DFS(root.left, sum) or self.DFS(root.right, sum)
else:
return False