leetcode-113 路径之和

leetcode-113 路径之和

题目描述：

给定一个二叉树和一个目标和，找到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。

解法一：找出所有路径，依次判断

# 注意函数名和变量名字冲突
class Solution:
    def pathSum(self, root: TreeNode, sum_: int) -> List[List[int]]:
        self.res = []
        self.findPath(root,[])
        return [v for v in self.res if sum(v)==sum_]
    
    
    def findPath(self,root,path):
        if not root:
            return
        if not root.left and not root.right:
            path.append(root.val)
            self.res.append(path)
            return 
        self.findPath(root.left,path+[root.val])
        self.findPath(root.right,path+[root.val])

解法二：直接判断，满足就加进去，速度比第一个慢

class Solution:
    def pathSum(self, root: TreeNode, sum_: int) -> List[List[int]]:
        self.res = []
        self.findPath(root,[],sum_)
        return self.res
    def findPath(self,root,path,sum_):
        if not root:
            return
        if not root.left and not root.right and root.val==sum_:
            # path.append(root.val)
            self.res.append(path+[root.val])
            return 
        self.findPath(root.left,path+[root.val],sum_-root.val)
        self.findPath(root.right,path+[root.val],sum_-root.val)

class Solution:
    def pathSum(self, root: TreeNode, sum: int) -> List[List[int]]:
        if not root:
            return []
        self.path = []
        self.tmp = sum
        
        def calSum(tmp_path,now,root):
            if not root:
                return False
            if not root.left and not root.right and root.val + now== self.tmp:
                self.path.append(tmp_path+[root.val])
            if root.left:
                calSum(tmp_path+[root.val],now+root.val,root.left)
            if root.right:
                calSum(tmp_path+[root.val],now+root.val,root.right)
        calSum([],0,root)
        return self.path