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feat: Implement binary tree path sum (#7135) #7270

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88 changes: 88 additions & 0 deletions data_structures/binary_tree/binary_tree_path_sum.py
Original file line number Diff line number Diff line change
@@ -0,0 +1,88 @@
"""
Given the root of a binary tree and an integer target,
find the number of paths where the sum of the values
along the path equals target.


Leetcode reference: https://leetcode.com/problems/path-sum-iii/
"""

from __future__ import annotations


class Node:
"""
A Node has value variable and pointers to Nodes to its left and right.
"""

def __init__(self, value: int) -> None:
self.value = value
self.left: Node | None = None
self.right: Node | None = None


class BinaryTreePathSum:
r"""
The below tree looks like this
10
/ \
5 -3
/ \ \
3 2 11
/ \ \
3 -2 1


>>> tree = Node(10)
>>> tree.left = Node(5)
>>> tree.right = Node(-3)
>>> tree.left.left = Node(3)
>>> tree.left.right = Node(2)
>>> tree.right.right = Node(11)
>>> tree.left.left.left = Node(3)
>>> tree.left.left.right = Node(-2)
>>> tree.left.right.right = Node(1)

>>> BinaryTreePathSum().path_sum(tree, 8)
3
>>> BinaryTreePathSum().path_sum(tree, 7)
2
>>> tree.right.right = Node(10)
>>> BinaryTreePathSum().path_sum(tree, 8)
2
"""

target: int

def __init__(self) -> None:
self.paths = 0

def dfs(self, node: Node | None, path_sum: int) -> None:
if node is None:
return

if path_sum == self.target:
self.paths += 1

if node.left:
self.dfs(node.left, path_sum + node.left.value)
if node.right:
self.dfs(node.right, path_sum + node.right.value)

def path_sum(self, node: Node | None, target: int | None = None) -> int:
if node is None:
return 0
if target is not None:
self.target = target

self.dfs(node, node.value)
self.path_sum(node.left)
self.path_sum(node.right)

return self.paths


if __name__ == "__main__":
import doctest

doctest.testmod()