[leetcode] 437. Path Sum III @ python

最新推荐文章于 2022-11-23 12:18:42 发布

原创最新推荐文章于 2022-11-23 12:18:42 发布 · 847 阅读

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本文探讨了在二叉树中寻找特定路径和的问题，通过深度优先搜索(DFS)和双重递归的方法，实现了在不从根节点或叶节点开始的情况下，查找所有向下路径的数目。算法的时间复杂度为O(n^2)，空间复杂度为O(1)，适用于1000个节点以内的树结构。

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原题

You are given a binary tree in which each node contains an integer value.

Find the number of paths that sum to a given value.

The path does not need to start or end at the root or a leaf, but it must go downwards (traveling only from parent nodes to child nodes).

The tree has no more than 1,000 nodes and the values are in the range -1,000,000 to 1,000,000.

Example:

root = [10,5,-3,3,2,null,11,3,-2,null,1], sum = 8

  10
 /  \
5   -3

/ \
3 2 11
/ \
3 -2 1

Return 3. The paths that sum to 8 are:

5 -> 3
5 -> 2 -> 1
-3 -> 11

解法

DFS + 双重递归. 定义dfs函数, 求在以当前节点能组成的path的个数, 有3种可能: 当前节点, 当前节点 + 左子节点的path, 当前节点 + 右子节点的path, 返回计数的个数. 然后在pathSum函数里, 再进行一次递归.

Time: O(n**2)
Space: O(1)

代码

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

class Solution:
    def pathSum(self, root: 'TreeNode', sum: 'int') -> 'int':
        if not root:
            return 0
        return self.dfs(root, sum) + self.pathSum(root.left, sum) + self.pathSum(root.right, sum)
        
    def dfs(self, root, sum):
        # count the number of paths starting from the node
        ans = 0
        if not root:
            return ans
        if root.val == sum:
            ans += 1
        ans += self.dfs(root.left, sum-root.val)
        ans += self.dfs(root.right, sum - root.val)
        return ans