112-path-sum

Given the root of a binary tree and an integer targetSum, return true if the tree has a root-to-leaf path such that adding up all the values along the path equals targetSum.

A leaf is a node with no children.

Example 1:

Input: root = [5,4,8,11,null,13,4,7,2,null,null,null,1], targetSum = 22 Output: true Explanation: The root-to-leaf path with the target sum is shown. Example 2:

Input: root = [1,2,3], targetSum = 5 Output: false Explanation: There are two root-to-leaf paths in the tree: (1 –> 2): The sum is 3. (1 –> 3): The sum is 4. There is no root-to-leaf path with sum = 5. Example 3:

Input: root = [], targetSum = 0 Output: false Explanation: Since the tree is empty, there are no root-to-leaf paths.

My solution：

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution(object):
    def hasPathSum(self, root, targetSum):
        """
        :type root: Optional[TreeNode]
        :type targetSum: int
        :rtype: bool
        """
        resList = []
        self.sumVal(root,targetSum,resList)
        return len(resList) >= 1
    def sumVal(self,root,target,resList):
        if not root:
            return
        if root.left:
            self.sumVal(root.left,target-root.val,resList)
        if root.right:
            self.sumVal(root.right,target-root.val,resList)
        if not root.left and not root.right:
            if target - root.val == 0:
                resList.append(target - root.val)
                return
            else:
                return