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描述
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:
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]]
Note:
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.
解析
根据题意，就是给出了一个二叉树，并且给出了一个需要删除的节点值列表 to_delete ，题目要求我们将节点值出现在 to_delete 中的节点都删去，并且将删除之后形成的多个森林的引用都放入列表中。
递归函数的操作是比较繁琐的，因为假如当前节点是需要被删除的，那么需要将其形成的左右非空两个子树加入到结果列表中，并且要将当前节点置空。这里的节点置空操作是需要注意的，因为结点置空不是直接 root=None ， 而是需要将其父节点的 left 引用或者 right 引用置空才对。另外需要注意的是别忘了将根节点也加入进结果列表中。
解答
class TreeNode(object):
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
class Solution(object):
def delNodes(self, root, to_delete):
"""
:type root: TreeNode
:type to_delete: List[int]
:rtype: List[TreeNode]
"""
result = []
def dfs(root):
if not root: return None
root.left = dfs(root.left)
root.right = dfs(root.right)
if root.val in to_delete:
if root.left:
result.append(root.left)
if root.right:
result.append(root.right)
return None
return root
dfs(root)
if root.val not in to_delete:
result.append(root)
return result
运行结果
Runtime: 110 ms, faster than 5.12% of Python online submissions for Delete Nodes And Return Forest.
Memory Usage: 13.9 MB, less than 92.47% of Python online submissions for Delete Nodes And Return Forest.
原题链接：leetcode.com/problems/de…
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