# Leetcode: Path Sum

Path Sum

Similar Problems:

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.

For 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.

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

# 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):
if root is None:
return False
queue = []
queue.append((root, 0))
while len(queue) != 0:
(element, current_sum) = queue[0]
del queue[0]
if element.left is None and element.right is None:
if element.val + current_sum == sum:
return True
if element.left:
queue.append((element.left, current_sum + element.val))
if element.right:
queue.append((element.right, current_sum + element.val))
return False

def hasPathSum_v1(self, root, sum):
"""
:type root: TreeNode
:type sum: int
:rtype: bool
"""
## Idea: DFS recursive, BFS
## Complexity: Time O(n), Space O(log(k))
return self._hasPathSum(root, sum, 0)

def _hasPathSum(self, root, sum, current_sum):
if root is None:
return False
if root.left is None and root.right is None:
return root.val + current_sum == sum

if root.left:
if self._hasPathSum(root.left, sum, current_sum + root.val):
return True

if root.right:
if self._hasPathSum(root.right, sum, current_sum + root.val):
return True

return False

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