题目:
python3 代码:
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def hasPathSum(self, root: TreeNode, sum: int) -> bool:
if not root:
return False
if not root.left and not root.right and sum == root.val:
return True
lefttree = self.hasPathSum(root.left, sum-root.val)
righttree = self.hasPathSum(root.right, sum-root.val)
if lefttree or righttree:
return True
else:
return False
思路:
使用递归想法,由于函数是返回bool类型的,所以只要将sum的值不断往下减,这样如果值减到0,这样就返回True, 就代表有Path 在BST里。否则,BST里是没有Path的。
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