動機
複習dfs
Problem
Given the root
of a binary tree and an integer targetSum
, return true
if the tree has a root-to-leaf path such that adding up all the values along the path equals targetSum
.
A leaf is a node with no children.
Example 1:
Input: root = [5,4,8,11,null,13,4,7,2,null,null,null,1], targetSum = 22Output: true
Example 2:
Input: root = [1,2,3], targetSum = 5Output: false
Example 3:
Input: root = [1,2], targetSum = 0Output: false
Constraints:
- The number of nodes in the tree is in the range
[0, 5000]
. -1000 <= Node.val <= 1000
-1000 <= targetSum <= 1000
Sol
注意到是root到leaf!!
def dfs(r,target,cnt):
if r is None:
return False
elif r.val+cnt == target and not r.left and not r.right:
return True
else:
return dfs(r.left,target,cnt+r.val) or dfs(r.right,target,cnt+r.val)
class Solution:
def hasPathSum(self, r: TreeNode, targetSum: int) -> bool:
return dfs(r,targetSum,0)