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.
Return the minimum number of operations needed to make the values at each level sorted in a strictly increasing order.
The level of a node is the number of edges along the path between it and the root node.
Example 1:
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.
Example 2:
Input: root = [1,3,2,7,6,5,4]
Output: 3
Explanation:
- Swap 3 and 2. The 2nd level becomes [2,3].
- Swap 7 and 4. The 3rd level becomes [4,6,5,7].
- Swap 6 and 5. The 3rd level becomes [4,5,6,7].
We used 3 operations so return 3.
It can be proven that 3 is the minimum number of operations needed.
Example 3:
Input: root = [1,2,3,4,5,6]
Output: 0
Explanation: Each level is already sorted in increasing order so return 0.
Constraints:
The number of nodes in the tree is in the range [1, 105].
1 <= Node.val <= 105
All the values of the tree are unique.
Solutions
Solution 1: BFS + Discretization + Element Swap
First, we traverse the binary tree using BFS to find the node values at each level. Then, we sort the node values at each level. If the sorted node values are different from the original node values, it means that we need to swap elements. The number of swaps is the number of operations needed at that level.
The time complexity is $O(n \times \log n)$. Here, $n$ is the number of nodes in the binary tree.
/** * Definition for a binary tree node. * type TreeNode struct { * Val int * Left *TreeNode * Right *TreeNode * } */funcminimumOperations(root*TreeNode)(ansint){f:=func(t[]int)int{varalls[]intfor_,v:=ranget{alls=append(alls,v)}sort.Ints(alls)m:=make(map[int]int)fori,v:=rangealls{m[v]=i}fori:=ranget{t[i]=m[t[i]]}res:=0fori:=ranget{fort[i]!=i{t[i],t[t[i]]=t[t[i]],t[i]res++}}returnres}q:=[]*TreeNode{root}forlen(q)>0{t:=[]int{}forn:=len(q);n>0;n--{node:=q[0]q=q[1:]t=append(t,node.Val)ifnode.Left!=nil{q=append(q,node.Left)}ifnode.Right!=nil{q=append(q,node.Right)}}ans+=f(t)}return}