[hard]450. Delete Node in a BST

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原创最新推荐文章于 2025-12-02 23:24:25 发布 · 273 阅读

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#leetcode #BST

本文介绍了一种在二叉搜索树(BST)中删除指定键值节点的方法。文章提供了详细的步骤说明，包括查找要删除的节点及之后的删除操作。文中还提供了一段C++代码示例，展示了如何实现这一过程。

450. 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:

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

Solution:

It reminds me of my high school days. It's a very basic step in treap, and now I just need to remove the fix from treap. Super easy.

1) If x has no sons, just delete it. don't forget x <- NULL

2) if x has a son, delete x and replace x with its son

2) if x has two son, right rotate x and deleteNode(x -> right)

(I assume you know what is a rotate operation in treap)

Here is my code:

class Solution {
public:
    TreeNode* deleteNode(TreeNode*& x, int key) {
        if (x) {
            if (x -> val == key) {
                if (x -> left == NULL && x -> right == NULL) {
                    delete x;
                    x = NULL;
                } else if (x -> left == NULL) {
                    auto y = x -> right;
                    delete x; x = y;
                } else if (x -> right == NULL) {
                    auto y = x -> left;
                    delete x; x = y;
                } else {
                    auto y = x -> left;
                    x -> left = y -> right;
                    y -> right = x;
                    x = y;
                    deleteNode(x -> right, key);
                }
            } else if (x -> val > key)
                deleteNode(x -> left, key);
            else
                deleteNode(x -> right, key);
        }
        return x;
    }
};