> For the complete documentation index, see [llms.txt](https://ondrej-kvasnovsky-2.gitbook.io/algorithms/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://ondrej-kvasnovsky-2.gitbook.io/algorithms/data-structures/binary-search-tree/delete-node-in-a-bst.md).

# 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
```

## Solution

According to the number of its children, we should consider three different cases ([video](https://www.youtube.com/watch?v=82cIlfCkCCw)):

1. If the target node has ***no child***, we can simply remove the node.
2. If the target node has ***one child***, we can use its child to replace itself.
3. If the target node has ***two children***, replace the node with its in-order successor or predecessor node and delete that node.

Here are examples of different cases to help you understand this strategy.![](https://s3-lc-upload.s3.amazonaws.com/uploads/2018/01/25/bst_deletion_case_1.png)![](https://s3-lc-upload.s3.amazonaws.com/uploads/2018/01/25/bst_deletion_case_2.png)\
![](https://s3-lc-upload.s3.amazonaws.com/uploads/2018/01/25/bst_deletion_case_3.png)

Iterative solution.

```
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    private TreeNode deleteRootNode(TreeNode root) {
        if (root == null) {
            return null;
        }
        if (root.left == null) {
            return root.right;
        }
        if (root.right == null) {
            return root.left;
        }
        TreeNode next = root.right;
        TreeNode pre = null;
        for(; next.left != null; pre = next, next = next.left);
        next.left = root.left;
        if(root.right != next) {
            pre.left = next.right;
            next.right = root.right;
        }
        return next;
    }

    public TreeNode deleteNode(TreeNode root, int key) {
        TreeNode cur = root;
        TreeNode pre = null;
        while(cur != null && cur.val != key) {
            pre = cur;
            if (key < cur.val) {
                cur = cur.left;
            } else if (key > cur.val) {
                cur = cur.right;
            }
        }
        if (pre == null) {
            return deleteRootNode(cur);
        }
        if (pre.left == cur) {
            pre.left = deleteRootNode(cur);
        } else {
            pre.right = deleteRootNode(cur);
        }
        return root;
    }
}
```

Recursive solution.

```
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    /**
     * findSuccesor - Helper function for finding successor
     * In BST, succesor of root is the leftmost child in root's right subtree
     */
    private TreeNode findSuccessor(TreeNode root) {
        TreeNode cur = root.right;
        while (cur != null && cur.left != null) {
            cur = cur.left;
        }
        return cur;
    }
    public TreeNode deleteNode(TreeNode root, int key) {
        // return null if root is null
        if (root == null) {
            return root;
        }

        // delete current node if root is the target node
        if (root.val == key) {
            // replace root with root->right if root->left is null    
            if (root.left == null) {
                return root.right;
            }

            // replace root with root->left if root->right is null
            if (root.right == null) {
                return root.left;
            }

            // replace root with its successor if root has two children
            TreeNode p = findSuccessor(root);
            root.val = p.val;
            root.right = deleteNode(root.right, p.val);
            return root;
        }
        if (root.val < key) {
            // find target in right subtree if root->val < key
            root.right = deleteNode(root.right, key);
        } else {
            // find target in left subtree if root->val > key
            root.left = deleteNode(root.left, key);
        }
        return root;
    }
}
```