LeetCode: 1325. Delete Leaves With a Given Value

题目

原题地址:https://leetcode.com/problems/delete-leaves-with-a-given-value/

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:

image1

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:

image2

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

Example 3:

image3

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].

解法

前序遍历二叉树,在遍历的过程中重建二叉树,将符合条件的节点删除(节点没有子节点并且节点的值等于 target )

这个方法的 Python 代码类似下面这样:

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution(object):
    def removeLeafNodes(self, root, target):
        if root is None:
            return

        if root.left is None and root.right is None and root.val == target:
            return

        root.left = self.removeLeafNodes(root.left, target)
        root.right = self.removeLeafNodes(root.right, target)

        # 应对 note 提到的 case:
        # 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).
        if root.left is None and root.right is None and root.val == target:
            return

        return root

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