标签：and The example bsp while exit turn wap remove
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:
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
class Solution { public: TreeNode* deleteNode(TreeNode* root, int key) { if(!root) return root; TreeNode* ret; if(root->val == key) { TreeNode* rnode_lmost = getlm_or_rm_node(root->right, true); if(rnode_lmost) { rnode_lmost->left = root->left; ret = root->right; }else ret = root->left; }else { if(key < root->val) root->left = deleteNode(root->left, key); else root->right = deleteNode(root->right, key); ret = root; } return ret; } TreeNode* getlm_or_rm_node(TreeNode* root, bool left){ if(!root) return root; if(left) { while(root->left) root = root->left; }else { while(root->right) root = root->right; } return root; } };
class Solution { public: TreeNode *deleteNode(TreeNode *root, int key) { TreeNode **cur = &root; while (*cur && (*cur)->val != key) cur = (key > (*cur)->val) ? &(*cur)->right : &(*cur)->left; if (*cur) { if (!(*cur)->right) *cur = (*cur)->left; else { TreeNode **successor = &(*cur)->right; while ((*successor)->left) successor = &(*successor)->left; swap((*cur)->val, (*successor)->val); *successor = (*successor)->right ? (*successor)->right : nullptr; } } return root; } };
标签：and The example bsp while exit turn wap remove
原文地址：https://www.cnblogs.com/ethanhong/p/10354389.html
