1 class Solution: 2 def __init__(self): 3 self.res = [] 4 self.temp = [] 5 6 def preNode(self,node,to_delete): 7 if node == None: 8 return None 9 val = node.val 10 if val in to_delete: 11 if node.left != None: 12 self.temp.append(node.left) 13 if node.right != None: 14 self.temp.append(node.right) 15 to_delete.remove(val) 16 return None 17 else: 18 if node.left != None: 19 leftval = node.left.val 20 if leftval in to_delete: 21 t = node.left 22 if t.left != None: 23 self.temp.append(t.left) 24 if t.right != None: 25 self.temp.append(t.right) 26 node.left = None 27 to_delete.remove(leftval) 28 else: 29 self.preNode(node.left,to_delete) 30 if node.right != None: 31 rightval = node.right.val 32 if rightval in to_delete: 33 t = node.right 34 if t.left != None: 35 self.temp.append(t.left) 36 if t.right != None: 37 self.temp.append(t.right) 38 node.right = None 39 to_delete.remove(rightval) 40 else: 41 self.preNode(node.right,to_delete) 42 return node 43 44 def delNodes(self, root: TreeNode, to_delete: 'List[int]') -> 'List[TreeNode]': 45 self.temp.append(root) 46 while len(to_delete) > 0 and len(self.temp) > 0: 47 t = self.temp.pop(0) 48 node = self.preNode(t,to_delete) 49 if node != None: 50 self.res.append(node) 51 self.res += self.temp 52 return self.res
思路：遍历二叉树，如果当前节点为被删除节点，则将其左右子树拆开成两个独立的树，返回空值。
如果左子树为被删除节点，则将左子树的左右子树拆开成两个独立的树，置当前节点的左子树为空，返回当前节点；否则，递归遍历左子树。
如果右子树为被删除节点，则将右子树的左右子树拆开成两个独立的树，置当前节点的右子树为空，返回当前节点；否则，递归遍历右子树。