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#!/usr/bin/env python3
###########################################################################
# script name: ch-2.py #
# #
# https://github.com/user-person #
# #
# https://perlweeklychallenge.org/blog/perl-weekly-challenge-056/ #
# #
# Path Sum #
# You are given a binary tree and a sum, write a script to find if #
# the tree has a path such that adding up all the values along the path #
# equals the given sum. Only complete paths (from root to leaf node) may #
# be considered for a sum. #
# #
# Example #
# Given the below binary tree and sum = 22, #
# #
# 5 #
# / \ #
# 4 8 #
# / / \ #
# 11 13 9 #
# / \ \ #
# 7 2 1 #
# #
# For the given binary tree, the partial path sum 5 -> 8 -> 9 = 22 is #
# not valid. #
# The script should return the path 5 -> 4 -> 11 -> 7 whose sum is 22 #
# #
###########################################################################
class Node:
def __init__(self,key):
self.left = None
self.right = None
self.val = key
branchMemory = []
k = 22
match = 0
def preOrder(root, toLeafSum):
if root:
toLeafSum += root.val
if root.left != None and root.right != None:
branchMemory.append(toLeafSum)
if root.left == None and root.right == None:
if toLeafSum == k:
global match
match += 1
if len(branchMemory) > 0:
toLeafSum = branchMemory.pop()
preOrder(root.left, toLeafSum)
preOrder(root.right, toLeafSum)
bTree = Node(5)
bTree.left = Node(4)
bTree.left.left = Node(11)
bTree.left.left.left = Node(7)
bTree.left.left.right = Node(2)
bTree.right = Node(8)
bTree.right.left = Node(13)
bTree.right.right = Node(9)
bTree.right.right.right = Node(1)
tLS = 0
preOrder(bTree,tLS)
if match == 0:
print('No branch equals', k)
elif match == 1:
print('1 branch equals', k)
else:
print(match, 'branches equal', k)
# output:
#
# 1 branch equals 22
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