[Oracle] LeetCode 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.

Solution

运用递归的方法来进行删除。首先根据 \(key\) 与当前节点的值，来判断是在左子树还是右子树。

当找到节点时，我们在右子树上不断找左子树的最后一个点，将这个点作为节点来补到原来的位置（因为这样才能满足 \(BST\)），然后在右子树上同样地递归来删除那个点

点击查看代码

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* deleteNode(TreeNode* root, int key) {
        if(!root)return root;
        if(root->val > key){
            root->left = deleteNode(root->left, key);
        }
        else if(root->val < key){
            root->right = deleteNode(root->right, key);
        }
        else{
            if(!root->right)return root->left;
            else if(!root->left)return root->right;
            else{
                TreeNode* tmp = root->right;
                while(tmp->left)tmp=tmp->left;
                root->val = tmp->val;
                root->right = deleteNode(root->right,tmp->val);
            }
        }
        return root;
    }
};