LC 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:

1. Search for a node to remove.
2. 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

 

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;
    }

};