Delete Node in a BST

LeetCode

第24天。

今天的题目是 Delete Node in a BST :

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:

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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

水题，只要先在BST上做搜索，然后删除就好了，因为只是BST，所以可以不考虑平衡的问题：

  • left和right都为空：直接删除，返回nullptr即可
  • left和right都不为空：默认采用把右子树的节点拉上来的方式，即把左子树插入到右子树中，然后再返回right即可。
  • left和right有一个不为空，则返回不为空的子树即可。

则代码如下：

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TreeNode* deleteNode(TreeNode *node) {
    auto left = node->left, right = node->right;
    delete node;
    if (left && right) {
        auto temp = right;
        while(temp->left) {
            temp = temp->left;
        }
        temp->left = left;
        return right;
    } 
    return (left ? left : (right ? right : nullptr));
}
TreeNode* deleteNode(TreeNode* root, int key) {
    if (root == nullptr) return nullptr;
    else if (root->val == key) {
        return deleteNode(root);
    } else if (key >  root->val) {
        root->right = deleteNode(root->right, key);    
    } else 
        root->left = deleteNode(root->left, key);
    
    return root;
}