# Get All Nodes On A Given Level In A Binary Tree With Python

## 04 September 2018 - 1 answer

i am trying to get a list of nodes (objetcs) in a python binary tree, i am looking for a recursive function implemented in the node object, so i will call function on the root node, and it will going down on childs nodes till reachs the specific level, and then will return those nodes in a list

My current aproach, i'm not sure if this is correct or the best way to implement it:

``````def get_level_nodes(self, nodes, level=1):
if self.level > level:
return nodes
if self.level == level:
nodes.append(self)
return nodes
for child in self.child_id:
nodes += child.get_level_nodes(node, level)
return nodes

# Getting the list
nodes_list = root_node.get_level_nodes([], 3)
``````

There is no real need to pass a list of nodes around. Each node can just return the appropriate level-nodes of its own subtree, and leave the combining of neighbours to the parent:

``````def get_level_nodes(self, level=1):
if self.level > level:
return []
if self.level == level:
return [self]
# child_id seems an odd name
return [n for c in self.children for n in c.get_level_nodes(level)]
``````

A more space-efficient implementation that does not build intermediate lists for each subtree would be a generator function:

``````def get_level_nodes(self, level=1):
if self.level > level:
return
if self.level == level:
yield self
else:
for c in self.children:
for n in c.get_level_nodes(level):
yield n
# or in Python3
# yield from c.get_level_nodes(level)

nodes_list = list(root_node.get_level_nodes(3))
``````