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113. Path Sum II

难度: Easy

刷题内容

原题连接

内容描述

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]
]

解题方案

思路 1 - 时间复杂度: O(N)- 空间复杂度: O(N)******

注意宁愿写几次curList + [root.val] 也不要直接传一个list进去,因为list pass by reference的亏已经吃过了

class Solution(object):
    def pathSum(self, root, sum):
        """
        :type root: TreeNode
        :type sum: int
        :rtype: List[List[int]]
        """
        res = []
        self.auxPathSum(root, sum, [], res)
        return res
    def auxPathSum(self, root, sum, cur_list, cur_lists):
        if not root:
            return
        sum -= root.val
        if sum == 0 and not root.left and not root.right:
            cur_lists.append(cur_list + [root.val])
            return 
        if root.left:
            self.auxPathSum(root.left, sum, cur_list + [root.val], cur_lists) 
        if root.right:
            self.auxPathSum(root.right, sum, cur_list + [root.val], cur_lists)