LeetCode-450. Delete Node in a BST

https://leetcode.com/problems/delete-node-in-a-bst/

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:

Search for a node to remove.
If the node is found, delete the node.
Note: Time complexity should be O(height of tree).

Example:

root = [5,3,6,2,4,null,7]
key = 3

    5
   / \
  3   6
 / \   \
2   4   7

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 following BST.

    5
   / \
  4   6
 /     \
2       7

Another valid answer is [5,2,6,null,4,null,7].

    5
   / \
  2   6
   \   \
    4   7

Solution(Python)

思路

1.首先递归的寻找要删除的节点。

2.找到要删除的节点后，有下面四种情况：

• 该节点左右子树均为空—返回null
• 该节点只有左子树为空—返回右子树
• 该节点只有右子树为空—返回左子树
• 该节点左右子树均不为空—寻找该节点的右子树中的最小节点，用最小节点的值替代该节点的值，继续递归地去删除右子树中的那个最小节点。

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def deleteNode(self, root, key):
        """
        :type root: TreeNode
        :type key: int
        :rtype: TreeNode
        """
        if not root:
            return None
        if key < root.val:
            root.left = self.deleteNode(root.left, key)
        elif key > root.val:
            root.right = self.deleteNode(root.right, key)
        else: # already found then delete
            if not root.left:
                return root.right
            elif not root.right:
                return root.left
            
            minNode = self.findMin(root.right)
            root.val = minNode.val
            root.right = self.deleteNode(root.right, root.val)
        
        return root
        
    def findMin(self, node):
        while node.left != None:
            node = node.left
        return node