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. 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

alt

Solution: Recursion¶

Time complexity: O(h)

Space complexity: O(h)

// Running time: 36 ms
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
  TreeNode* deleteNode(TreeNode* root, int key) {
    if (root == nullptr) return root;
    if (key > root->val) {
      root->right = deleteNode(root->right, key);
    } else if (key < root->val) {
      root->left = deleteNode(root->left, key);
    } else {
      TreeNode* new_root = nullptr;
      if (root->left == nullptr) {
        new_root = root->right;
      } else if (root->right == nullptr) {
        new_root = root->left;
      } else {
        // Find the min node in the right sub tree
        TreeNode* parent = root;
        new_root = root->right;
        while (new_root->left != nullptr) {
          parent = new_root;
          new_root = new_root->left;
        }

        if (parent != root) {
          parent->left = new_root->right;
          new_root->right = root->right;
        }

        new_root->left = root->left;
      }

      delete root;
      return new_root;
    }

    return root;
  }
};

v2

// Running time: 35 ms
class Solution {
public:
  TreeNode* deleteNode(TreeNode* root, int key) {
    if (root == nullptr) return root;
    if (key > root->val) {
      root->right = deleteNode(root->right, key);
    } else if (key < root->val) {
      root->left = deleteNode(root->left, key);
    } else {
      if (root->left != nullptr && root->right != nullptr) {
        TreeNode* min = root->right;
        while (min->left != nullptr) min = min->left;
        root->val = min->val;
        root->right = deleteNode(root->right, min->val);
      } else {
        TreeNode* new_root = root->left == nullptr ? root->right : root->left;
        root->left = root->right = nullptr;
        delete root;
        return new_root;
      }
    }
    return root;
  }
};