1110. Delete Nodes And Return ⭐ Forest

Medium
Given the root of a binary tree, each node in the tree has a distinct value.
After deleting all nodes with a value in to_delete, we are left with a forest (a disjoint union of trees).
Return the roots of the trees in the remaining forest. You may return the result in any order.
Example 1:
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Input: root = [1,2,3,4,5,6,7], to_delete = [3,5]
Output: [[1,2,null,4],[6],[7]]
Example 2:
Input: root = [1,2,4,null,3], to_delete = [3]
Output: [[1,2,4]]
Constraints:
  • The number of nodes in the given tree is at most 1000.
  • Each node has a distinct value between 1 and 1000.
  • to_delete.length <= 1000
  • to_delete contains distinct values between 1 and 1000.

解題

/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func delNodes(root *TreeNode, to_delete []int) []*TreeNode {
m := make(map[int]bool)
for _, n := range to_delete {
m[n] = true
}
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ans := make([]*TreeNode, 0)
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var dfs func(*TreeNode, bool) *TreeNode
dfs = func(r *TreeNode, isRoot bool) *TreeNode {
if r == nil { return nil }
if isRoot && !m[r.Val] {
ans = append(ans, r)
}
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r.Left = dfs(r.Left, m[r.Val])
r.Right = dfs(r.Right, m[r.Val])
​
if m[r.Val] { return nil }
return r
}
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dfs(root, true)
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return ans
}
​