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].
analysis
The meaning of the question is: delete the leaf node whose node value is equal to the target. If after deleting the node, its parent node has no leaf node and its value is equal to the target, it needs to be deleted. The idea of recursion is also very straightforward. After traversal, if the values of the current left and right nodes are empty, and the current node is target, delete and return None, otherwise return to the current node. I took a look at other people's implementations, and the idea was similar to mine.
Code
# 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