【leetcode】1325. Delete Leaves With a Given Value

题目如下：

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
• Each tree has at most 3000 nodes.
• Each node's value is between [1, 1000].

解题思路：最直接的方法就是循环做删除特定值的叶子节点的操作，只能没有符合条件的叶子节点为止。

代码如下：

# 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,parent,target,direction):
        if node.left is None and node.right is None and node.val == target:
            self.flag = True
            if direction == 'L':
                parent.left = None
            else:
                parent.right = None
        if node.left != None:
            self.recursive(node.left,node,target,'L')

        if node.right != None:
            self.recursive(node.right,node,target,'R')



    def removeLeafNodes(self, root, target):
        """
        :type root: TreeNode
        :type target: int
        :rtype: TreeNode
        """
        while True:
            self.flag = False
            if root.left == None and root.right == None and root.val == target:
                return None
            node = root
            self.recursive(node,None,target,'')
            if self.flag == False:
                break
        return root