Path Sum III

Definition for a Binary Tree Node

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

Solution

Vanilla DFS

Complexity

time: $O(n \log n)$ (worst case if tree is a line: $O(n^2)$)
space: $O(\log n)$ (due to implicit stack space, worst case: $O(n)$)

def pathSum(self, root: TreeNode, sum: int) -> int:
    if not root:
        return 0

    def dfs(node: TreeNode, csum: int) -> int:
        if not node:
            return 0
        csum += node.val
        return (csum == sum) + dfs(node.left, csum) + dfs(node.right, csum)

    return dfs(root, 0) + self.pathSum(root.left, sum) + self.pathSum(root.right, sum)

Recursive DFS w/ Memoization

Say the current node has a path sum csum, if there exists a node that had been visited and had a path sum csum - sum, then increment res by memo[csum - sum].

As a node's subtree is being traversed, increment memo[csum] by $1$.

Complexity

time: $O(n)$
space: $O(n)$

from collections import defaultdict
def pathSum(self, root: TreeNode, sum: int) -> int:
    res = 0
    memo = defaultdict(int)
    memo[0] = 1

    def dfs(node, csum):
        nonlocal res
        if not node:
            return
        csum += node.val
        # increment by the # of times
        res += memo[csum - sum]
        # mark the current node as visited
        memo[csum] += 1
        dfs(node.left, csum)
        dfs(node.right, csum)
        memo[csum] -= 1

    dfs(root, 0)
    return res