LeetCode 437. Path Sum III

本文介绍了一种在二叉树中寻找路径和等于指定值的方法。该方法不局限于从根节点到叶节点的路径,而是可以跨越任意节点,并向下遍历。通过前缀和的技巧,算法能在O(n)的时间复杂度内解决问题。

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Problem Statement

(Source) 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

Tags: Tree.

Solution

Idea: TODO.

# 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 pathSum(self, root, sum):
        """
        :type root: TreeNode
        :type sum: int
        :rtype: int
        """
        prefix_sum = {0 : 1}
        return self.helper(root, sum, prefix_sum, 0)

    def helper(self, root, target, prefix_sum, cur):
        if not root: return 0
        total = cur + root.val
        res = prefix_sum.get(total - target, 0)
        prefix_sum[total] = prefix_sum.get(total, 0) + 1
        res += self.helper(root.left, target, prefix_sum, total) + \
                self.helper(root.right, target, prefix_sum, total)
        prefix_sum[total] -= 1
        return res

Complexity Analysis:

  • Time Complexity: O(n) .
  • Space Complexity: ? .
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