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原题
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.
解法
递归, base case有3种情况: 1) root为空时, 返回False 2) root没有子树且root的值等于sum, 返回True 3) root没有子树且root的值不等于sum, 返回False. 然后再从root的左右子树中寻找结果.
Time: O(n) , n为节点的数量
Space: O(1)
代码
# 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 not root:
return False
if root.left is None and root.right is None and root.val == sum:
return True
if root.left is None and root.right is None and root.val != sum:
return False
return self.hasPathSum(root.left, sum - root.val) or self.hasPathSum(root.right, sum - root.val)