Leetcode #919: Complete Binary Tree Inserter
In this guide, we solve Leetcode #919 Complete Binary Tree Inserter in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
A complete binary tree is a binary tree in which every level, except possibly the last, is completely filled, and all nodes are as far left as possible. Design an algorithm to insert a new node to a complete binary tree keeping it complete after the insertion.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Tree, Breadth-First Search, Design, Binary Tree
Intuition
We need level-by-level exploration or shortest steps, which is ideal for BFS.
A queue naturally models the frontier of the search.
Approach
Push initial nodes into a queue and expand in layers.
Track visited nodes to prevent cycles.
Steps:
- Initialize queue with start nodes.
- Process level by level.
- Track visited nodes.
Example
Input
["CBTInserter", "insert", "insert", "get_root"]
[[[1, 2]], [3], [4], []]
Output
[null, 1, 2, [1, 2, 3, 4]]
Explanation
CBTInserter cBTInserter = new CBTInserter([1, 2]);
cBTInserter.insert(3); // return 1
cBTInserter.insert(4); // return 2
cBTInserter.get_root(); // return [1, 2, 3, 4]
Python Solution
# 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
class CBTInserter:
def __init__(self, root: Optional[TreeNode]):
self.tree = []
q = deque([root])
while q:
for _ in range(len(q)):
node = q.popleft()
self.tree.append(node)
if node.left:
q.append(node.left)
if node.right:
q.append(node.right)
def insert(self, val: int) -> int:
p = self.tree[(len(self.tree) - 1) // 2]
node = TreeNode(val)
self.tree.append(node)
if p.left is None:
p.left = node
else:
p.right = node
return p.val
def get_root(self) -> Optional[TreeNode]:
return self.tree[0]
# Your CBTInserter object will be instantiated and called as such:
# obj = CBTInserter(root)
# param_1 = obj.insert(val)
# param_2 = obj.get_root()
Complexity
The time complexity is O(V+E). The space complexity is , where is the number of nodes in the tree.
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.