leetcode 450. Delete Node in a BST-优快云博客

本文链接：https://blog.youkuaiyun.com/liqinzhe11/article/details/78955067

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。文章详细阐述了如何定位目标节点并进行删除操作，同时保持树的结构不变。提供了一种高效的算法实现，时间复杂度为O(tree height)。

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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

这个题写了很久！真的最近有点生疏了。

1、涉及删除节点，需要知道删除节点的父节点才好操作。

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)
    {
        if (!root) return NULL;
        else if (root->val == key && !root->left && !root->right)   return nullptr;
        else if (root->val == key && root->left && !root->right)   return root->left;
        else if (root->val == key && !root->left && root->right)   return root->right;
        
        TreeNode* q = root;
        TreeNode* parent = NULL;    //涉及删除节点，需要知道这个删除节点的父节点。
        
        while (q)   //定位key值点的位置
        {
            if (q->val > key)
            {
                parent = q;
                q = q->left;
            }
            else if (q->val < key)
            {
                parent = q;
                q = q->right;
            }
            else   //q就是被删除的点
            {
                if (!q->left)
                {
                    if (q == parent->left)   parent->left = q->right;
                    else                     parent->right = q->right;
                    delete q;
                }
                else   //找到左子树的最大值，并且删除那个节点
                {
                    q->val = deleteMaxPoint(q->left, q);
                }
                break;
            }
        }
        return root;
    }
    
    int deleteMaxPoint (TreeNode * root, TreeNode *parent)   //找到root节点的下面的（包括root节点）最大值，并且删除那个节点
    {
        if (!root->right)  //BST没有右节点，说明root节点的值最大
        {
            int ret = root->val;
            if (root == parent->left)   parent->left = root->left;
            else                     parent->right = root->left;
            delete root;
            return ret;
        }
        return deleteMaxPoint(root->right, root);
    }
};