Leetcode 450.删除二叉搜索树中的结点

原题连接：

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.

Example 1:

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 following BST.
Please notice that another valid answer is [5,2,6,null,4,null,7] and its also accepted.

Example 2:

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:

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

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

做法一：递归

BST的左右子树也都是BST，因此常考虑用递归来解题

思路：

空树就返回空 根据BST性质，比root小就递归左子树，比root大就递归右子树 如果root就是key：

如果没有左右孩子，直接删除 只有左孩子或右孩子，删除root，并用左孩子或右孩子代替root的位置 左右孩子都有，则找到比root大的最小节点successor（右子树的最左边），在右子树中将successor删除后，让successor来代替root的位置

c++代码：

/**
 * 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) {
          
   
        // 如果是空树, 返回空
        if(root == nullptr)
            return nullptr;

        // 比根节点值小, 递归左子树
        if(key < root->val){
          
   
            root->left = deleteNode(root->left, key);
            return root;
        }

        // 比根节点值大, 递归右子树
        if(key > root->val){
          
   
            root->right = deleteNode(root->right, key);
            return root;
        }

        // 删除结点
        if(key == root->val){
          
   
            // 如果没有左右孩子
            if(!root->left && !root->right){
          
   
                return nullptr;
            }

            // 只有左孩子, 没有右孩子
            if (!root->right) {
          
   
                return root->left;
            }

            // 只有右孩子, 没有左孩子
            if (!root->left) {
          
   
                return root->right;
            }

            else{
          
   
                // 左右孩子都有, 找到比root大的最小结点successor(在右子树的最左边)
                TreeNode *successor = root->right;
                while (successor->left) {
          
   
                    successor = successor->left;
                }

                // 在root的右子树中删除successor
                root->right = deleteNode(root->right, successor->val);

                // 用successor来代替root的位置
                successor->right = root->right;
                successor->left = root->left;
                return successor;
            }
        }
        return root;
    }
};

复杂度分析

时间复杂度：O(n)，n为节点总数，最坏情况下，找successor和删除它都需要便利整个树 空间复杂度：O(n)，n为节点总数，递归的最大深度就是n