Leetcode #2471: Minimum Number of Operations to Sort a Binary Tree by Level
In this guide, we solve Leetcode #2471 Minimum Number of Operations to Sort a Binary Tree by Level 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
You are given the root of a binary tree with unique values. In one operation, you can choose any two nodes at the same level and swap their values.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Tree, Breadth-First Search, 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: root = [1,4,3,7,6,8,5,null,null,null,null,9,null,10]
Output: 3
Explanation:
- Swap 4 and 3. The 2nd level becomes [3,4].
- Swap 7 and 5. The 3rd level becomes [5,6,8,7].
- Swap 8 and 7. The 3rd level becomes [5,6,7,8].
We used 3 operations so return 3.
It can be proven that 3 is the minimum number of operations needed.
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 Solution:
def minimumOperations(self, root: Optional[TreeNode]) -> int:
def swap(arr, i, j):
arr[i], arr[j] = arr[j], arr[i]
def f(t):
n = len(t)
m = {v: i for i, v in enumerate(sorted(t))}
for i in range(n):
t[i] = m[t[i]]
ans = 0
for i in range(n):
while t[i] != i:
swap(t, i, t[i])
ans += 1
return ans
q = deque([root])
ans = 0
while q:
t = []
for _ in range(len(q)):
node = q.popleft()
t.append(node.val)
if node.left:
q.append(node.left)
if node.right:
q.append(node.right)
ans += f(t)
return ans
Complexity
The time complexity is . The space complexity is O(V).
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.