# Find number of neighbour pairs in the input tree

My code works with small binary tree but with big ones (around 10000 nodes) it takes 15 sec to process.

For small binary tree I'm always getting correct answer and it is fast.

How can I make my program faster for big binary tree?

(x, y) is a neighbour pair when the depth of x and y is the same, the node color of x and y is the same, the node key of x and y are equal and there is no other node of the same node color between x and y on the horizontal line connecting x and y in standard drawing..

### Input

The first line of input contains number of nodes and root node. Next, each line describes each node.

The first value is the node label, the second value is the node key, the third and the fourth values represent the labels of the left and right child respectively, and the fifth value represents the node color, white is 0, black is 1. If any of the children does not exist there is value 0 instead of the child label at the corresponding place

### Output

number of neighbour pairs in the input tree

### My code

from collections import defaultdict
inputs = inputstring.split(" ")
nodes_number = int(inputs[0])
root_node = int(inputs[1])
input_tree = [list(map(int, input().split())) for _ in range(nodes_number)]
tree = {}
result = {}
def depth(node, count):
for i in input_tree:
node_label = i[0]
if node_label == node:
tree.setdefault(count, []).append(i)
l_node, r_node = i[2], i[3]
if l_node != 0 and r_node != 0:
depth(l_node, count+1)
depth(r_node, count+1)

elif l_node != 0:
depth(l_node, count+1)

elif r_node != 0:
depth(r_node, count+1)
depth(root_node, 0)

def neighbour_pair(input_tree):
for value in input_tree.values():
for i in range(len(value)):
for j in range(len(value)):
if j > i:
if value[i][1] == value[j][1] and value[i][4] == value[j][4]:
result.setdefault(value[i][0], []).append(value[j][0])
neighbour_pair(tree)
return len(result)
if __name__ == '__main__':


### Example of small input

13 5
7 50 0 0 1
2 70 10 11 0
4 30 9 0 0
6 70 0 0 0
1 90 8 12 0
9 90 0 2 1
13 90 0 6 0
5 30 4 3 0
12 80 0 0 1
10 50 0 0 1
11 50 0 0 0
3 80 1 13 0
8 70 7 0 1


Its associated representation:

The labels are written in the yellow rectangle at the corresponding nodes. Their key is the number in the black or white oval. For this example, the neighbour pairs are (1, 13), (2, 6), (10, 7).

As such, the output of the program for this tree is:

3

• Welcome to Code Review! I hope you get some good answers. – Phrancis Apr 25 '16 at 22:19

Overall, conventions are respected: case, spaces… But there are still some things missing to make your code feel readable:

• vertical spacing: you should add blank lines before functions definition and important logical sections of the code so we can grasp the structure at a glance.
• naming: one letter variable are not great, especially when conventions dictates an other use of that letter than the one you make (I’m looking at you for i in input_tree:).
• namedtuples can help give meaning to your lines of input:

from collections import namedtuple

Node = namedtuple('Node', 'id key left_child right_child color')

…
input_tree = [Node(*map(int, input().split())) for _ in range(nodes_number)]
…
for node in input_tree:
node_label = node.id
…
l_node, r_node = node.left_child, node.right_child


and so on.

• You don't need to define nested functions, you can pass them parameters. It will be easier to test (for correctness or speed). In fact, you should separate further the computation bits from the parsing of the input.

You also import defaultdict but never use it. Instead, your code contains a few setdefault(…, []) on regular dictionaries. You should declare tree = defaultdict(list) and result = defaultdict(list) instead.

# Building the depth of nodes

For each node (technically each node id), you iterate over the whole input_tree and then recurse, iterating over the whole input_tree for the two children.

This is a huge waste of time. Instead of iterating over a list to find a node by its id, you should use a datastructure that let you access a node by its id directly. A simple dictionary where keys are ids and values are node should suffice.

# Computing neighbour pairs

First of, you could start your last loop at i+1 so you don't need the if j > i: test. But still, you don't even need that second loop at all: by keeping track of the last white node and the last black node you could only iterate once for each depth.

# Proposed improvements

from collections import defaultdict, namedtuple

Node = namedtuple('Node', 'id key left_child right_child color')

for _ in range(nodes_count):
yield Node(*map(int, input().split()))

def parse_input():
nodes_count, root_node = map(int, input().split())
tree = {node.id: node for node in read_nodes(nodes_count)}
return root_node, tree

def build_depth(node_id, tree, storage, current_depth=0):
node = tree[node_id]
storage[current_depth].append(node)
for child in (node.left_child, node.right_child):
if child:
build_depth(child, tree, storage, current_depth + 1)

def neighbour_pairs(layered_tree):
neighbours_count = 0 # neighbours = defaultdict(list)
for nodes in layered_tree.values():
last_black = Node(0, None, 0, 0, 1)
last_white = Node(0, None, 0, 0, 0)
for node in nodes:
if node.color == 1:
last_node, last_black = last_black, node
else:
last_node, last_white = last_white, node
if last_node.key == node.key:
neighbours_count += 1 # neighbours[last_node.id] = node.id
return neighbours_count # return neighbours

if __name__ == '__main__':
root, tree = parse_input()
nodes_by_depth = defaultdict(list)
build_depth(root, tree, nodes_by_depth)
neighbours = neighbour_pair(nodes_by_depth)
print(neighbours) # print(len(neighbours))


You can see that I’m not storing the pairs of neighbours, it will help speed things up. But I left comments if you still want to be able to retrieve them to test for correctness.