【leetcode】1110. Delete Nodes And Return Forest

題目如下：

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]]

 

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.

解題思路：從根節點開始，判斷是否在to_delete，如果不在，把這個節點加入Output中，往左右子樹方向繼續遍歷；如果在，把其左右子節點加入queue中；而后從queue中依次讀取元素，並對其做與根節點一樣的操作，直到queue為空位置。

代碼如下：

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def recursive(self,node,queue,to_delete):
        if node.left != None and node.left.val in to_delete:
            queue.append(node.left)
            node.left = None
        if node.right != None and node.right.val in to_delete:
            queue.append(node.right)
            node.right = None
        if node.left != None:
            self.recursive(node.left,queue,to_delete)
        if node.right != None:
            self.recursive(node.right,queue,to_delete)
    def delNodes(self, root, to_delete):
        """
        :type root: TreeNode
        :type to_delete: List[int]
        :rtype: List[TreeNode]
        """
        if root == None:
            return []
        queue = [root]
        res = []
        while len(queue) > 0:
            node = queue.pop(0)
            if node.val not in to_delete:
                res.append(node)
                self.recursive(node,queue,to_delete)
            else:
                if node.left != None:
                    queue.append(node.left)
                if node.right != None:
                    queue.append(node.right)

        return res