LeetCode 450 Delete Node in a BST（删除BST节点）

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:

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

题目大意：删除BST上的节点。

代码如下：

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     struct TreeNode *left; *     struct TreeNode *right; * }; */ struct TreeNode* findMax(struct TreeNode* root){    if(root == NULL) return NULL;    else if(root->right == NULL) return root;    else return findMax(root->right);}struct TreeNode* deleteNode(struct TreeNode* root, int key) {    if(root == NULL) return NULL;    if(root->val < key){        root->right = deleteNode(root->right, key);    }else if(root->val > key){        root->left = deleteNode(root->left, key);    }else if(root->left && root->right){        struct TreeNode* tmp = findMax(root->left);        root->val = tmp->val;        root->left = deleteNode(root->left, tmp->val);    }else{        struct TreeNode* tmp = root;        if(root->left == NULL) root = root->right;        else if(root->right == NULL) root = root->left;        free(tmp);    }    return root;}