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
思路：
1.若要删除的结点不在树中，直接返回根结点
2.找到结点有下面四种情况
- 该节点是叶子结点
- 该结点只有左结点
- 该结点只有右结点
- 该结点有两个结点，找到该结点右子树的最小结点，把该节点的值替换为最小结点的值，删除最小结点
1 class Solution { 2 public: 3 bool find(TreeNode* root, int key){ 4 if(root == NULL) return false; 5 if(key < root->val) return find(root->left, key); 6 else if(key > root->val) return find(root->right, key); 7 else return true; 8 } 9 10 TreeNode* deleteNode(TreeNode* root, int key) { 11 if(!find(root, key)) return root; 12 if(key < root->val) root->left = deleteNode(root->left, key); 13 else if(key > root->val) root->right = deleteNode(root->right, key); 14 else{ 15 if(root->left == NULL && root->right == NULL) root = NULL; 16 else if(root->left != NULL && root->right == NULL) root = root->left; 17 else if(root->right != NULL && root->left == NULL) root = root->right; 18 else{ 19 TreeNode *temp = root->right; 20 while(temp->left != NULL) temp = temp->left; 21 root->val = temp->val; 22 root->right = deleteNode(root->right, temp->val); 23 } 24 } 25 return root; 26 } 27 };