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

 

Approach #1: C++.[recursive]

/**
 * 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 nullptr;
        if (root->val < key) root->right = deleteNode(root->right, key);
        else if (root->val > key) root->left = deleteNode(root->left, key);
        else {
            if (root->left == nullptr) return root->right;
            else if (root->right == nullptr) return root->left;
            else {
                TreeNode* minNode = findMinNode(root->right);
                root->val = minNode->val;
                root->right = deleteNode(root->right, root->val);
            }
        }
        
        return root;
    }
    
private:   
    TreeNode* findMinNode(TreeNode* node) {
        if (node->left != nullptr) findMinNode(node->left);
        return node;
    }
};

　　

Approach #2: Java.[recursive]

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode deleteNode(TreeNode root, int key) {
        if (root == null) return null;
        if (root.val > key) root.left = deleteNode(root.left, key);
        else if (root.val < key) root.right = deleteNode(root.right, key);
        else {
            if (root.left == null)
                return root.right;
            else if (root.right == null)
                return root.left;
            else {
                TreeNode minNode = findMinNode(root.right);
                root.val = minNode.val;
                root.right = deleteNode(root.right, root.val);
            }
        }
        return root;
    }
    
    private TreeNode findMinNode (TreeNode root) {
        while (root.left != null) {
            root = root.left;
        }
        return root;
    }
}

　　

Appraoch #3: Python.[Iterator]

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None



class Solution(object):
    def deleteNode(self, root, key):
        """
        :type root: TreeNode
        :type key: int
        :rtype: TreeNode
        """
        cur = root
        pre = None
        while cur is not None and cur.val is not key:
            pre = cur
            if key < cur.val:
                cur = cur.left
            elif key > cur.val:
                cur = cur.right
        if pre is None:
            return self.deleteRootNode(cur)
        if pre.left == cur:
            pre.left = self.deleteRootNode(cur)
        else:
            pre.right = self.deleteRootNode(cur)        
        return root
    
    def deleteRootNode(self, node):
        if node is None:
            return None
        if node.left is None:
            return node.right
        if node.right is None:
            return node.left

        nextNode = node.right
        pre = None
        while nextNode.left is not None:
            pre = nextNode
            nextNode = nextNode.left
        nextNode.left = node.left
        if node.right is not nextNode:
            pre.left = nextNode.right
            nextNode.right = node.right
        return nextNode