LeetCode450. Delete Node in a BST

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文章目录

- 一、题目
- 二、题解

一、题目

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.

Example 1:

Input: root = [5,3,6,2,4,null,7], key = 3
Output: [5,4,6,2,null,null,7]
Explanation: 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 above BST.
Please notice that another valid answer is [5,2,6,null,4,null,7] and it’s also accepted.

Example 2:

Input: root = [5,3,6,2,4,null,7], key = 0
Output: [5,3,6,2,4,null,7]
Explanation: The tree does not contain a node with value = 0.
Example 3:

Input: root = [], key = 0
Output: []

Constraints:

The number of nodes in the tree is in the range [0, 104].
-105 <= Node.val <= 105
Each node has a unique value.
root is a valid binary search tree.
-105 <= key <= 105

Follow up: Could you solve it with time complexity O(height of tree)?

二、题解

/*** 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 nullptr;else if(root->val == key){//叶子节点if(!root->left && !root->right) return nullptr;//左为空右不为空else if(!root->left && root->right) return root->right;//左不为空右为空else if(root->left && !root->right) return root->left;//左右均不为空else if(root->left && root->right){TreeNode* cur = root->right;while(cur->left) cur = cur->left;cur->left = root->left;return root->right;}}else if(root->val > key)root->left = deleteNode(root->left,key);else if(root->val < key)root->right = deleteNode(root->right,key);return root;}
};