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path-sum.py
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38 lines (26 loc) · 894 Bytes
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'''
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.
'''
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def hasPathSum(self, root: TreeNode, sum: int) -> bool:
# Approach one
if not root: return False
if sum == root.val and not root.left and not root.right: return True
return self.hasPathSum(root.left , sum - root.val) or self.hasPathSum(root.right , sum - root.val)