112-113. Path Sum I & II [Easy & Medium] 递归和DFS

112. Path Sum

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
class Solution(object):
    def hasPathSum(self, root, sum):
        """
        :type root: TreeNode
        :type sum: int
        :rtype: bool
        """
        if root == None:
            return False
        if root.left == None and root.right == None and root.val == sum:
            return True
        res = False
        if root.left:
            res = self.hasPathSum(root.left, sum-root.val)
        if res:
            return True
        if root.right:
            res = self.hasPathSum(root.right, sum-root.val)
        return res

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113. Path Sum II

113. Path Sum II

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def pathSum(self, root, sum):
        """
        :type root: TreeNode
        :type sum: int
        :rtype: List[List[int]]
        """
        res = []
        trace = []
        self.DFS(root, sum, res, trace, 0)
        return res
    def DFS(self, root, sum, res, trace, now):
        if root == None:
            return
        trace.append(root.val)
        if root.val == sum and root.left == None and root.right == None:
            res.append(trace[:])
            trace.pop()
            return
        if root.left:
            self.DFS(root.left, sum-root.val, res, trace, now)
        if root.right:
            self.DFS(root.right, sum-root.val, res, trace, now)
        trace.pop()