LeetCode: 437. Path Sum III

题目

原题地址:https://leetcode.com/problems/path-sum-iii/

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:

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

解法

先限定 path 只能从 root 节点开始,这样的话, 按照 112. Path Sum 的思路, 相应的 Python 代码类似下面这样:

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution(object):
    def __init__(self):
        self._count = 0

    def pathSum(self, root, target_sum):

        self._preorder(root, target_sum)

        return self._count

    def _preorder(self, root, target_sum):
        if root is None:
            return 0

        if root.val == target_sum:
            self._count += 1

        new_sum = target_sum - root.val

        self._preorder(root.left, new_sum)
        self._preorder(root.right, new_sum)

但是,因为题目中说了不仅限于从 root 开始: 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). 所以,需要修改上面的代码,让每个节点都走一遍上面从 root 开始的查找过程。

这个思路的 Python 代码类似下面这样:

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution(object):
    def __init__(self):
        self._count = 0

    def pathSum(self, root, target_sum):
        if root is None:
            return 0

        self._preorder(root, target_sum)

        self.pathSum(root.left, target_sum)
        self.pathSum(root.right, target_sum)

        return self._count

    def _preorder(self, root, target_sum):
        if root is None:
            return 0

        if root.val == target_sum:
            self._count += 1

        new_sum = target_sum - root.val

        self._preorder(root.left, new_sum)
        self._preorder(root.right, new_sum)

Comments

comments powered by Disqus