113. Path Sum II

Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.

Note: A leaf is a node with no children.

Example:

Given the below binary tree and sum = 22,

      5
     / 
    4   8
   /   / 
  11  13  4
 /      / 
7    2  5   1

Return:

[
   [5,4,11,2],
   [5,8,4,5]
]
# 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 __init__(self):
        self.res = []
        
    def pathSum(self, root, sum):
        """
        :type root: TreeNode
        :type sum: int
        :rtype: List[List[int]]
        """
        if root is None:
             return []
        self.helper(root, sum, [])
        return self.res
    def helper(self, root, sum, tmp_l):
        if root.left is None and root.right is None:
            if sum == root.val:
                self.res.append(tmp_l + [root.val])
            else:
                return None
        #每次进入函数,取剩下的值,并且重新创建节点值列表
        if root.left:
            self.helper(root.left, sum-root.val, tmp_l+[root.val])
        if root.right:
            self.helper(root.right, sum-root.val, tmp_l+[root.val])
            
            
原文地址:https://www.cnblogs.com/boluo007/p/12441236.html