動機

找右邊的最小，之後重建tree

Problem

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:

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

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

 

Example 1:

Input: root = [5,3,6,2,4,Leetcode,7], key = 3Output: [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:

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

Example 3:

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

 

Constraints:

  • The number of nodes in the tree is in the range [0, 104].
  • -105 <= Node.val <= 105
  • Each node has a unique value.
  • root is a valid binary search tree.
  • -105 <= key <= 105

Sol

class Solution:
    def deleteNode(self, root: Optional[TreeNode], key: int) -> Optional[TreeNode]:
        if not root:
            return None
        elif root.val > key:
            root.left = self.deleteNode(root.left, key)
            return root
        elif root.val < key:
            root.right = self.deleteNode(root.right, key)
            return root
        else:
            if not root.left and not root.right:
                return None
            elif not root.left:
                return root.right
            elif not root.right:
                return root.left
            else:
                # find min in root.right
                now = root.right
                prev = root
                while now.left:
                    prev, now = now, now.left
                
                if now == root.right: # no left
                    now.left = root.left
                else:
                    prev.left = now.right
                    now.left = root.left
                    now.right = root.right
                return now