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].
分析
题目的意思是：删除结点值等于target的叶子结点，如果删除了结点之后，其父结点没有叶子结点并且并且其值等于target，则还需要删除。递归的思路也很直接，后序遍历，然后如果当前的左右结点的值都为空，并且当前的结点为target，则删除返回None，否则返回当前的结点。我看了一下别人的实现，思路跟我差不多。
代码
# 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:
if(root is None):
return root
root.left=self.removeLeafNodes(root.left,target)
root.right=self.removeLeafNodes(root.right,target)
if(root.val==target):
if(not root.left and not root.right):
return None
return root