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

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本文介绍了一种从二叉搜索树（BST）中删除指定键的方法。文章详细解释了搜索要删除节点的过程，并给出了如何通过调整子节点来保持BST性质的代码实现。

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;
}