LeetCode 2471 - Minimum Number of Operations to Sort a Binary Tree by Level

Difficulty: 🟡 Medium
Topics: Tree, Breadth-First Search, Binary Tree

Solution

Problem Understanding

The problem asks us to determine the minimum number of swap operations required to sort the values of a binary tree level by level in strictly increasing order. Specifically, at each level of the tree, we can only swap values of nodes that exist on that level. The tree is given as a standard binary tree with unique integer values, and the goal is to count the minimum number of such swaps for all levels combined.

The input is the root of a binary tree, which can contain up to 100,000 nodes, each with a unique integer value. The output is a single integer representing the total minimum operations needed. Because node values are unique, we do not need to handle duplicates, simplifying sorting and mapping operations.

Key edge cases include trees that are already level-sorted (no operations needed), trees with only one node, and trees with multiple levels requiring complex swaps. The problem guarantees a valid binary tree with unique values, so we do not have to handle invalid or empty inputs beyond the minimal size of one node.

Approaches

Brute Force

A brute-force approach would iterate through each level, repeatedly swapping pairs of nodes until the level is sorted. This could involve nested loops comparing all pairs of nodes and swapping them as needed. While correct, this is highly inefficient: for a level with n nodes, we could have up to O(n^2) operations per level, and the total number of nodes is up to 100,000. This approach would be far too slow for large trees.

Optimal Approach

The key insight is that the minimum number of swaps to sort an array can be calculated efficiently by finding the number of cycles in the permutation that maps the current array to its sorted version. Specifically, for an array of length n, the number of swaps required to sort it is sum(len(cycle) - 1) over all cycles in the permutation.

For our problem, the optimal solution consists of these steps:

Perform a level-order traversal (BFS) of the tree.
For each level, collect the node values into a list.
Determine the minimum number of swaps to sort that list using the cycle detection method.
Sum the swaps for all levels.

This approach reduces the problem to sorting arrays and counting cycles, which is efficient and scales to the problem constraints.

Approach	Time Complexity	Space Complexity	Notes
Brute Force	O(N^2)	O(N)	Swap repeatedly until each level is sorted, inefficient for large levels
Optimal	O(N log N)	O(N)	BFS to collect levels + cycle detection per level, scales to 10^5 nodes

Algorithm Walkthrough

Initialize a queue for BFS traversal and add the root node.
Initialize a variable total_swaps to track operations.
While the queue is not empty, process each level:
Determine the number of nodes at this level (level_size).
Collect the values of all nodes at this level in a list.
Generate a mapping from value to its index in the sorted version of the list.
Use a visited array to track nodes already part of cycles.
Iterate over nodes in the level list:
If a node is not visited, traverse its cycle.
Count the cycle length and add cycle_length - 1 to the level swap count.
Add the level swap count to total_swaps.
Enqueue all children of nodes at this level into the queue.
Return total_swaps.

Why it works: Each level is treated independently, and the minimum number of swaps for an array is correctly calculated by counting cycles. Since levels are independent and node values are unique, the sum of swaps across all levels gives the correct minimum total operations.

Python Solution

from collections import deque
from typing import Optional

# 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:
        if not root:
            return 0
        
        total_swaps = 0
        queue = deque([root])
        
        while queue:
            level_size = len(queue)
            level_vals = []
            
            for _ in range(level_size):
                node = queue.popleft()
                level_vals.append(node.val)
                if node.left:
                    queue.append(node.left)
                if node.right:
                    queue.append(node.right)
            
            # Compute minimum swaps to sort this level
            sorted_vals = sorted(level_vals)
            index_map = {val: i for i, val in enumerate(level_vals)}
            visited = [False] * level_size
            level_swaps = 0
            
            for i in range(level_size):
                if visited[i] or index_map[sorted_vals[i]] == i:
                    continue
                
                cycle_size = 0
                j = i
                while not visited[j]:
                    visited[j] = True
                    j = index_map[sorted_vals[j]]
                    cycle_size += 1
                level_swaps += cycle_size - 1
            
            total_swaps += level_swaps
        
        return total_swaps

The implementation first performs BFS to traverse levels. For each level, it collects node values and determines the minimum number of swaps using cycle detection. This ensures each level is sorted with the fewest operations.

Go Solution

package main

type TreeNode struct {
    Val   int
    Left  *TreeNode
    Right *TreeNode
}

func minimumOperations(root *TreeNode) int {
    if root == nil {
        return 0
    }

    totalSwaps := 0
    queue := []*TreeNode{root}

    for len(queue) > 0 {
        levelSize := len(queue)
        levelVals := make([]int, levelSize)
        
        for i := 0; i < levelSize; i++ {
            node := queue[0]
            queue = queue[1:]
            levelVals[i] = node.Val
            if node.Left != nil {
                queue = append(queue, node.Left)
            }
            if node.Right != nil {
                queue = append(queue, node.Right)
            }
        }

        sortedVals := make([]int, levelSize)
        copy(sortedVals, levelVals)
        sort.Ints(sortedVals)
        
        indexMap := make(map[int]int)
        for i, val := range levelVals {
            indexMap[val] = i
        }

        visited := make([]bool, levelSize)
        levelSwaps := 0

        for i := 0; i < levelSize; i++ {
            if visited[i] || indexMap[sortedVals[i]] == i {
                continue
            }
            cycleSize := 0
            j := i
            for !visited[j] {
                visited[j] = true
                j = indexMap[sortedVals[j]]
                cycleSize++
            }
            levelSwaps += cycleSize - 1
        }

        totalSwaps += levelSwaps
    }

    return totalSwaps
}

The Go implementation mirrors the Python logic closely. The key differences include handling slices instead of lists and using maps instead of dictionaries. The BFS and cycle detection logic remains identical.

Worked Examples

Example 1

Input: [1,4,3,7,6,8,5,null,null,null,null,9,null,10]

Level	Values	Sorted	Cycles	Swaps
0	[1]	[1]	0	0
1	[4,3]	[3,4]	1	1
2	[7,6,8,5]	[5,6,7,8]	2	2
3	[9,10]	[9,10]	0	0

Total swaps = 1 + 2 = 3

Complexity Analysis

Measure	Complexity	Explanation
Time	O(N log N)	BFS traverses all nodes O(N), sorting each level can take O(k log k) for level size k, total nodes sum to N
Space	O(N)	Queue for BFS can store up to O(N) nodes at the last level, plus arrays for level values

The BFS ensures we touch each node once, and cycle detection is linear per level. Sorting dominates the runtime, giving O(N log N) overall.

Test Cases

# Provided examples
assert Solution().minimumOperations(TreeNode(1, TreeNode(4, TreeNode(7), TreeNode(6)), TreeNode(3, TreeNode(8), TreeNode(5, TreeNode(9), TreeNode(10))))) == 3
assert Solution().minimumOperations(TreeNode(1, TreeNode(3, TreeNode(7), TreeNode(6)), TreeNode(2, TreeNode(5), TreeNode(4)))) == 3
assert Solution().minimumOperations(TreeNode(1, TreeNode(2), TreeNode(3, TreeNode(4), TreeNode(5, TreeNode(6))))) == 0

# Edge case: single node
assert