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# 450     Delete Node in a BST

## 450. [Delete Node in a BST](https://leetcode.com/problems/delete-node-in-a-bst/description/)

## 1. Question

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
```

## 2. Implementation

**(1) Iteration**

```java
class Solution {
    public TreeNode deleteNode(TreeNode root, int key) {
        TreeNode preNode = null, curNode = root;

        while (curNode != null && curNode.val != key) {
            preNode = curNode;
            if (curNode.val < key) {
                curNode = curNode.right;
            }
            else{
                curNode = curNode.left;
            }
        }

        if (preNode == null) {
            return getSuccessor(curNode);
        }
        else if (preNode.left == curNode) {
            preNode.left = getSuccessor(curNode);
        }
        else if (preNode.right == curNode) {
            preNode.right = getSuccessor(curNode);
        }
        return root;
    }

    public TreeNode getSuccessor(TreeNode node) {
        if (node == null) {
            return null;
        }

        if (node.left == null) {
            return node.right;
        }

        if (node.right == null) {
            return node.left;
        }

        TreeNode preNode = null, successor = node.right;

        while (successor.left != null) {
            preNode = successor;
            successor = successor.left;
        }

        successor.left = node.left;
        if (node.right != successor) {
            preNode.left = successor.right;
            successor.right = node.right;
        }
        return successor;
    }
}
```

## 3. Time & Space Complexity

**Iteration:**&#x65F6;间复杂度O(h), 空间复杂度O(1)