LeetCode 450. Delete Node in a BST

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题目

Given a root node reference of a BST and a key, delete the node with the given key in the BST. Return the root node reference (possibly updated) of the BST.

Basically, the deletion can be divided into two stages:

  1. Search for a node to remove.
  2. If the node is found, delete the node.

Follow up: Can you solve it with time complexity O(height of tree)?

Example 1:

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Input: root = [5,3,6,2,4,null,7], key = 3
Output: [5,4,6,2,null,null,7]
Explanation: Given key to delete is 3. So we find the node with value 3 and delete it.
One valid answer is [5,4,6,2,null,null,7], shown in the above BST.
Please notice that another valid answer is [5,2,6,null,4,null,7] and it's also accepted.

Example 2:

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Input: root = [5,3,6,2,4,null,7], key = 0
Output: [5,3,6,2,4,null,7]
Explanation: The tree does not contain a node with value = 0.

Example 3:

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Input: root = [], key = 0
Output: []

Constraints:

  • The number of nodes in the tree is in the range [0, 10^4].
  • -10^5 <= Node.val <= 10^5
  • Each node has a unique value.
  • root is a valid binary search tree.
  • -10^5 <= key <= 10^5

思路

  1. 被删除节点没有左子树：返回其右子树
  2. 被删除节点节点没有右子树：返回其左子树
  3. 被删除节点既有左子树，又有右子树：
    1）查找到其右子树的最小值的节点，替换掉被删除的节点，并删除找到的最小节点
    2）查找到其左子树的最大值的节点，替换掉被删除的节点，并删除找到的最大节点

下面的做法是查找右子树的最小值节点的方法，最小节点就是右子树中的最靠左边的节点。代码使用的递归，最核心的是找到该节点之后的操作，特别是把值进行交换一步很重要，因为我们并没有删除了该最小值节点，所以把最小值的节点赋值成要查找的节点，然后在之后的操作中将会把它删除。

代码

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# 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 deleteNode(self, root: TreeNode, key: int) -> TreeNode:
        if not root: return None
        if root.val == key:
            if not root.right:
                left = root.left
                return left
            else:
                right = root.right
                while right.left:
                    right = right.left
                root.val, right.val = right.val, root.val
        root.left = self.deleteNode(root.left, key)
        root.right = self.deleteNode(root.right, key)
        return root