Question 42 of 75 - LeetCode Weekly Challenge

LeetCode Challenge #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.


Example 1:

del node 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.
del node supp

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: []



  • 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
Video Solution
C++ Solution
					class Solution {
    TreeNode* nodeDeleted(TreeNode* root, int key)
        if(root==NULL)return root;
        else if(key>root->val)root->right=nodeDeleted(root->right,key);
            if(!root->left and !root->right)return NULL;
            else if(!root->left||!root->right) return root->left?root->left:root->right;

            TreeNode* temp= root->right;
        return root;
    TreeNode* deleteNode(TreeNode* root, int key) {
        return nodeDeleted(root,key);

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