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Delete Node in a BST LeetCode Solution

Problem – 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:

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.

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

Delete Node in a BST LeetCode Solution in Java

public TreeNode deleteNode(TreeNode root, int key) {
    if(root == null){
        return null;
    if(key < root.val){
        root.left = deleteNode(root.left, key);
    }else if(key > root.val){
        root.right = deleteNode(root.right, key);
        if(root.left == null){
            return root.right;
        }else if(root.right == null){
            return root.left;
        TreeNode minNode = findMin(root.right);
        root.val = minNode.val;
        root.right = deleteNode(root.right, root.val);
    return root;

private TreeNode findMin(TreeNode node){
    while(node.left != null){
        node = node.left;
    return node;

Delete Node in a BST LeetCode Solution in C++

class Solution {
    TreeNode* deleteNode(TreeNode* root, int key) {
            if(key < root->val) root->left = deleteNode(root->left, key);     //We frecursively call the function until we find the target node
            else if(key > root->val) root->right = deleteNode(root->right, key);       
                if(!root->left && !root->right) return NULL;          //No child condition
                if (!root->left || !root->right)
                    return root->left ? root->left : root->right;    //One child contion -> replace the node with it's child
					                                                //Two child condition   
                TreeNode* temp = root->left;                        //(or) TreeNode *temp = root->right;
                while(temp->right != NULL) temp = temp->right;     //      while(temp->left != NULL) temp = temp->left;
                root->val = temp->val;                            //       root->val = temp->val;
                root->left = deleteNode(root->left, temp->val);  //        root->right = deleteNode(root->right, temp);		
        return root;

Delete Node in a BST LeetCode Solution in Python

class Solution(object):
    def deleteNode(self, root, key):
        if not root: return None
        if root.val == key:
            if not root.right: return root.left
            if not root.left: return root.right
            if root.left and root.right:
                temp = root.right
                while temp.left: temp = temp.left
                root.val = temp.val
                root.right = self.deleteNode(root.right, root.val)

        elif root.val > key:
            root.left = self.deleteNode(root.left, key)
            root.right = self.deleteNode(root.right, key)
        return root
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