LeetCode

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.

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

查找并删除BST中某个结点。

这道题，在我写了96行代码还在wa后，对你没看错，96行！我果断放弃了这个代码，围观评论区大神orzzzzzzz

然后发现自己蠢的没救了。。。为啥非要执着于把那个结点删除然后调个结点过来呢，咱就不能直接改变结点值么然后再去删除想要调的那个结点么！！！震惊哭QAQ

递归简直是脑残人的救星，在不会做的情况下，我还是不要纠结非得写非递归吧，找块豆腐撞死算了。

一路搜索下去，沿途改变各个结点的左右状态，一直都搜索到key结点，找到key的右子树中值最小的结点，赋值给key，并删除这个结点。时间复杂度O(h)

/** * Definition for a binary tree node. * 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) return root;        if (key < root->val)            root->left = deleteNode(root->left, key);        else if (key > root->val)            root->right = deleteNode(root->right, key);        else {            if (!root->left)                 return root->right;            if (!root->right)                return root->left;                        TreeNode* cur = finMin(root->right);            root->val = cur->val;            root->right = deleteNode(root->right, cur->val);        }        return root;    }    TreeNode* finMin(TreeNode* root) {        while (root->left)            root = root->left;        return root;    }};