解题思路

只需要分类讨论就行，根据不同情况进行不同的处理

  1. 节点为空：return root
  2. 节点==val：
    1. 左右不为空：将要删除的节点的左树连接到右树的最左节点的左
    2. 左为空，右不为空：return root.right
    3. 左不为空，右为空：return root.left
  3. 节点>val：说明val在当前节点的左子树中
  4. 节点<val：说明val在当前节点的右子树中

代码

Python

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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: Optional[TreeNode], key: int) -> Optional[TreeNode]:
        if root is None: return root
        if root.val == key:
            if root.left and root.right:
                cur = root.right
                # 找到右孩子的最左叶子节点
                while cur.left:
                    cur = cur.left
                cur.left = root.left
                return root.right
            elif not root.right: return root.left
            elif not root.left: return root.right
            else: return None
        if root.val > key: root.left = self.deleteNode(root.left, key)
        if root.val < key: root.right = self.deleteNode(root.right, key)
        return root

C++

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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* deleteNode(TreeNode* root, int key) {
        if (root == NULL) return root;
        if (root->val == key) {
            if (root->left == NULL) return root->right;
            else if (root->right == NULL) return root->left;
            else {
                TreeNode* cur = root->right;
                while (cur->left != NULL) {
                    cur = cur->left;
                }
                cur->left = root->left;
                TreeNode* tmp = root;
                root = root->right;
                delete tmp;
                return root;
            }
        }
        else if (root->val > key) root->left = deleteNode(root->left, key);
        else if (root->val < key) root->right = deleteNode(root->right, key);
        return root;
    }
};

Go

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func deleteNode(root *TreeNode, key int) *TreeNode {
    if (root == nil) {return root}
    if (root.Val == key) {
        if (root.Left != nil && root.Right != nil) {
            cur := root.Right
            for cur.Left != nil {
                cur = cur.Left
            }
            cur.Left = root.Left
            return root.Right
        } else if (root.Left == nil) {
            return root.Right
        } else if (root.Right == nil) {
            return root.Left
        }
    } else if (root.Val > key) {
        root.Left = deleteNode(root.Left, key)
    } else if (root.Val < key) {
        root.Right = deleteNode(root.Right, key)
    }
    return root

}