# Path Sum III

TreeDFS

## # 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

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## # 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)

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### # 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

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