112PathSum.py

"""
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.
"""
"""
Comments
"""
"""
My
"""
# 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:
        if root is None:
            return False
        if not (root.left or root.right): # 已经到达叶节点了
            return sum == root.val
        return self.hasPathSum(root.left, sum-root.val) or self.hasPathSum(root.right, sum-root.val)
"""
Fast
"""
class Solution:
    def hasPathSum(self, root: 'TreeNode', sum: 'int') -> 'bool':
        node = root
        return self.traverse(node,0,sum)
    
    def traverse(self,node,s,sum):
        if node == None:
            return False
        s += node.val
        if node.left == None and node.right == None:
            return s == sum
        return self.traverse(node.left,s,sum) or self.traverse(node.right,s,sum)