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.
LeetCode:链接
这题和剑指Offer_编程题24:二叉树中和为某一值的路径(DFS)形式差不多,但是要简单。
用递归的思想。如果根节点的左右子树都为空,那么返回根的值是否等于sum,否者,sum 减去根的值得到一个newsum,判断newsum在左子树或者右子树是否有路径。
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
def hasPathSum(self, root, sum):
"""
:type root: TreeNode
:type sum: int
:rtype: bool
"""
if root == None:
return False
if root.left == None and root.right == None and root.val == sum:
return True
return self.hasPathSum(root.left, sum-root.val) or self.hasPathSum(root.right, sum-root.val)