LeetCode PathSum

112 Path Sum 和113 Path Sum2 关注的是从root到leaf的情况，437 Path Sum3 关注的是任一点 downwards到某一个点的情况。 三题感觉都是比较典型的dfs题。

112 
# 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, s: int) -> bool:
        if not root: return False 
        
        if root and not root.right and not root.left: return root.val == s 
        
        return self.hasPathSum(root.left, s-root.val) or self.hasPathSum(root.right, s-root.val)

113
# 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: TreeNode, s: int) -> List[List[int]]:
        if not root: return []
        
        self.res = []
        
        self.dfs(root, s, [], 0)
        
        return self.res 
    
    def dfs(self, root, s, path, cur):
        if not root:
            return 
        if cur + root.val== s and not root.left and not root.right:
          
            path+=[root.val]
            
            self.res.append(path)
            return 
        self.dfs(root.left, s, path+[root.val], cur+root.val)
        self.dfs(root.right, s, path+[root.val], cur+root.val)

437
# 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: TreeNode, s: int) -> int:
        if not root: return 0
        
        return (self.dfs(root, s, 0, [0])[0]//2 + self.pathSum(root.left, s) + self.pathSum(root.right, s))
      
    def dfs(self, root, s, cur, count):
        if cur == s:
            count[0] += 1 
        if not root:
            return count 
        self.dfs(root.left, s, cur+root.val, count)
        self.dfs(root.right, s, cur+root.val, count)
        
        return count