0112. Path Sum.md

题目

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.

我的思路

这道题用递归可以轻易求解，递归返回True的条件是sum==当前节点的val。若当前节点不存在，则说明该路径不符合要求，返回False。若当前节点存在，则令sum=sum-val，然后继续递归。

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def hasPathSum(self, root: TreeNode, sum: int) -> bool:
        if not root:
            return False
        if not root.left and not root.right:
            if sum == root.val:
                return True
        if root.left or root.right:
            sum = sum - root.val
        return self.hasPathSum(root.left,sum) or self.hasPathSum(root.right,sum)