LeetCode 437. Path Sum III 时间复杂度（O(n^2))

路径总和的动态规划解法

最新推荐文章于 2024-09-24 13:34:48 发布

ziyue246

最新推荐文章于 2024-09-24 13:34:48 发布

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本文介绍了一种使用动态规划和后序遍历解决二叉树路径总和问题的方法，通过递归构建字典记录节点到目标和的距离，最终统计所有满足条件的路径数量。

时间复杂度（O(n^2)，思想：后序遍历+动态规划

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



class Solution:
    count = 0
    def update_dict(self, dict1, dict2, value, targe_sum):
        for key in dict1:
            curr_value = key + value
            if curr_value not in dict2:
                dict2[curr_value] = dict1[key]
            else:
                dict2[curr_value] += dict1[key]
        dict1_value = targe_sum-value
        if dict1_value in dict1:
            self.count += dict1[dict1_value]

    def pathCount(self, root: TreeNode, target_sum: int) -> dict():
        if root == None: return {}

        curr_left_map = self.pathCount(root.left, target_sum)
        curr_right_map = self.pathCount(root.right, target_sum)
        curr_map2 = {}
        self.update_dict({0: 1}, curr_map2, root.val, target_sum)
        self.update_dict(curr_left_map, curr_map2, root.val, target_sum)
        self.update_dict(curr_right_map, curr_map2, root.val, target_sum)
        del curr_left_map, curr_right_map
        return curr_map2

    def pathSum(self, root: TreeNode, sum: int) -> int:
        self.pathCount(root, sum)
        return self.count