【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】
二叉树剪枝问题， 同样是递归的方式来做， 但是相比之前的常规遍历问题来讲， 会用一个变量借助每层递归的返回值。
比如此题中，每层递归都有对应的if判定，一部分if用于return终结递归，一部分用变量接住return回的值。
类似题型：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: # 符合删除条件
return None
else:
return root
return trim(root)