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112. Path Sum.py
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112. Path Sum.py
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# 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.
# Note: A leaf is a node with no children.
# return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.
# Runtime: 32 ms, faster than 72.74% of Python online submissions for Path Sum.
# Memory Usage: 16.3 MB, less than 6.82% of Python online submissions for Path Sum.
# 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
"""
# Return False if root is empty
if root is None:
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)
# Shorter Solution
# if root is None:
# return False
# if not root.left and not root.right:
# return root.val == sum
# return self.hasPathSum(root.left, sum - root.val) or self.hasPathSum(root.right, sum - root.val)