leetcode 450. Delete Node in a BST 二叉搜索树BST删除结点

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

题意很简单，就是考察二叉搜索树BST的结点删除，记住吧，这是十分经典的问题

代码如下：

#include <iostream>#include <vector>#include <map>#include <set>#include <queue>#include <stack>#include <string>#include <climits>#include <algorithm>#include <sstream>#include <functional>#include <bitset>#include <cmath>using namespace std;/*struct TreeNode {     int val;     TreeNode *left;     TreeNode *right;     TreeNode(int x) : val(x), left(NULL), right(NULL) {}};*/class Solution {public:    TreeNode* deleteNode(TreeNode* root, int key)     {        if (root == NULL)            return root;        else        {            if (root->val == key)            {                if (root->right == NULL)                {                    TreeNode* left = root->left;                    delete root;                    return left;                }                else                {                    TreeNode* right = root->right;                    while (right->left != NULL)                        right = right->left;                    swap(root->val,right->val);                }            }            root->left = deleteNode(root->left, key);            root->right = deleteNode(root->right, key);            return root;        }    }};