path-sum-iii

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Description

Given the root of a binary tree and an integer targetSum, return the number of paths where the sum of the values along the path equals targetSum.

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

 

Example 1:

Input: root = [10,5,-3,3,2,null,11,3,-2,null,1], targetSum = 8
Output: 3
Explanation: The paths that sum to 8 are shown.

Example 2:

Input: root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22
Output: 3

 

Constraints:

• The number of nodes in the tree is in the range [0, 1000].
• -109 <= Node.val <= 109
• -1000 <= targetSum <= 1000

Solution(Python)

class Solution(object):
    def pathSum(self, root, sum):
        def dfs(sumHash, prefixSum, node):

            if not node:
                return 0

            # Sum of current path
            prefixSum += node.val

            # number of paths that ends at current node
            path = sumHash[prefixSum - sum]

            # add currentSum to prefixSum Hash
            sumHash[prefixSum] += 1

            # traverse left and right of tree
            path += dfs(sumHash, prefixSum, node.left) + dfs(
                sumHash, prefixSum, node.right
            )

            # remove currentSum from prefixSum Hash
            sumHash[prefixSum] -= 1

            return path

        # depth first search, initialize sumHash with prefix sum of 0, occurring once
        return dfs(collections.defaultdict(int, {0: 1}), 0, root)