450. Delete Node in a BST

最新推荐文章于 2021-01-27 20:57:34 发布

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最新推荐文章于 2021-01-27 20:57:34 发布

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分类专栏： LeetCode LeetCode 文章标签： LeetCode BST delete

本文链接：https://blog.youkuaiyun.com/teffi/article/details/53898215

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本文介绍了一种在二叉搜索树(BST)中删除指定节点的方法，详细讲解了两种情况下的删除步骤，并提供了LeetCode上的AC代码实现。一种情况是当节点只有一个或没有子节点时的操作，另一种情况涉及两个子节点时如何寻找后继节点并进行替换。

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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

解答如下：
1. 如果待删除节点最多只有一个孩子，孩子节点替换待删除节点，删除节点。
2. 如果待删除节点有两个孩子节点，找到右子树中最小的节点（后继节点），待删除节点的值替换为后继节点值，继续在右子树中删除后继节点。

LeetCode AC代码如下：

/**
 * 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) {
        TreeNode *temp;
        if (root != NULL) {
            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 != NULL && root->right != NULL) {
                temp = minNode(root->right);
                if (temp != NULL)  {
                    root->val = temp->val;
                    root->right = deleteNode(root->right, root->val);
                }
            }
            else {
                temp = root;
                if (root->left == NULL) root = root->right;
                else if (root->right == NULL)  root = root->left;
                delete temp;
            }
        }
        return root;
    }

    TreeNode *minNode(TreeNode *root) {
        TreeNode *pNode = root;
        while (pNode != NULL && pNode->left != NULL) {
            pNode = pNode->left;
        }
        return pNode;
    }
};