leetcodeday113 –路径总和 II

给你二叉树的根节点 root 和一个整数目标和 targetSum ，找出所有 从根节点到叶子节点 路径总和等于给定目标和的路径。

叶子节点 是指没有子节点的节点。

示例 1：

输入：root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22
输出：[[5,4,11,2],[5,8,4,5]]

示例 2：

输入：root = [1,2,3], targetSum = 5
输出：[]

示例 3：

输入：root = [1,2], targetSum = 0
输出：[]

# [113] 路径总和 II
#

# @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 pathSum(self, root: Optional[TreeNode], targetSum: int) -> List[List[int]]:
        res=list()
        rev=list()
        
        def subisBalanced(root,deep,target,rev):
            if  root.left==None and root.right==None and deep==target:   
                res.append(rev)
                return 
            if root.left!=None:
                deepl=deep+ root.left.val
                
                subisBalanced(root.left,deepl,target,rev+ [root.left.val])

                
            if  root.right!=None:
                deepr=deep+root.right.val
               
                subisBalanced(root.right,deepr,target,rev+[root.right.val])

        if root==None:
            return []
        else: 
            rev.append(root.val)
            subisBalanced(root,root.val, targetSum,rev)
            return res
        
# @lc code=end