leetcodeday112–路径总和

给你二叉树的根节点 root 和一个表示目标和的整数 targetSum 。判断该树中是否存在 根节点到叶子节点 的路径，这条路径上所有节点值相加等于目标和 targetSum 。如果存在，返回 true ；否则，返回 false 。

叶子节点 是指没有子节点的节点。

#
# @lc app=leetcode.cn id=112 lang=python3
#
# [112] 路径总和
#

# @lc code=start
# 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: Optional[TreeNode], targetSum: int) -> bool:
        def subisBalanced(root,deep,target):
            if root.left!=None:
                deepl=deep+root.left.val
                if subisBalanced(root.left,deepl,target):
                    return True            
            if  root.right!=None:
                deepr=deep+root.right.val 
                if subisBalanced(root.right,deepr,target):
                    return True
            
            if  root.left==None and root.right==None and deep==target:    
                return True
        if root==None:
            return False
        else:
            if subisBalanced(root,root.val,targetSum)==True:
                return True
            else:
                return False

# @lc code=end