【Description】
Given a binary tree root and an integer target, delete all the leaf nodes with value target.

Note that once you delete a leaf node with value target, if it's parent node becomes a leaf node and has the value target, it should also be deleted (you need to continue doing that until you can't).

Example 1:

Input: root = [1,2,3,2,null,2,4], target = 2
Output: [1,null,3,null,4]
Explanation: Leaf nodes in green with value (target = 2) are removed (Picture in left).
After removing, new nodes become leaf nodes with value (target = 2) (Picture in center).
Example 2:

Input: root = [1,3,3,3,2], target = 3
Output: [1,3,null,null,2]
Example 3:

Input: root = [1,2,null,2,null,2], target = 2
Output: [1]
Explanation: Leaf nodes in green with value (target = 2) are removed at each step.
Example 4:

Input: root = [1,1,1], target = 1
Output: []
Example 5:

Input: root = [1,2,3], target = 1
Output: [1,2,3]

Constraints:

1 <= target <= 1000
The given binary tree will have between 1 and 3000 nodes.
Each node's value is between [1, 1000].

【Idea】
二叉樹剪枝問題， 同樣是遞歸的方式來做， 但是相比之前的常規(guī)遍歷問題來講， 會用一個(gè)變量借助每層遞歸的返回值。
比如此題中，每層遞歸都有對應(yīng)的if判定，一部分if用于return終結(jié)遞歸，一部分用變量接住return回的值。
類似題型：814， 669

【Solution】

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def removeLeafNodes(self, root: TreeNode, target: int) -> TreeNode:
        def trim(root):
            if not root: 
                return root 
            if root.left:
                root.left = trim(root.left)
            if root.right:
                root.right = trim(root.right)
            if root.val == target and not root.left and not root.right:    # 符合刪除條件
                return None
            else:
                return root 
                    
        return trim(root)