Formatted question description: https://leetcode.ca/all/450.html

450. Delete Node in a BST (Medium)

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

Companies:
Microsoft, Google, Amazon

Related Topics:
Tree

Solution 1.

// OJ: https://leetcode.com/problems/delete-node-in-a-bst/
// Time: O(H)
// Space: O(H)
class Solution {
public:
    TreeNode* deleteNode(TreeNode* root, int key) {
        if (!root) 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) {
            auto p = root->left;
            while (p->right) p = p->right;
            root->val = p->val;
            root->left = deleteNode(root->left, root->val);
        } else if (root->right) {
            auto p = root->right;
            while (p->left) p = p->left;
            root->val = p->val;
            root->right = deleteNode(root->right, root->val);
        } else {
            delete root;
            root = NULL;
        }
        return root;
    }
};

Solution 2.

// OJ: https://leetcode.com/problems/delete-node-in-a-bst/
// Time: O(H)
// Space: O(H)
class Solution {
public:
    TreeNode* deleteNode(TreeNode* root, int key) {
        if (!root) 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) {
            auto right = root->right;
            delete root;
            return right;
        } else if (!root->right) {
            auto left = root->left;
            delete root;
            return left;
        } else {
            auto node = root->right;
            while (node->left) node = node->left;
            root->val = node->val;
            root->right = deleteNode(root->right, root->val);
        }
        return root;
    }
};

Java

  • /**
     * 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 root;
            if (root.val == key) {
                TreeNode dummyRoot = new TreeNode(0);
                dummyRoot.left = root;
                if (root.right != null) {
                    TreeNode successor = root.right;
                    if (successor.left == null) {
                        successor.left = root.left;
                        dummyRoot.left = successor;
                    } else {
                        TreeNode parent = successor, child = successor.left;
                        while (child.left != null) {
                            child = child.left;
                            parent = parent.left;
                        }
                        root.val = child.val;
                        parent.left = child.right;
                    }
                } else
                    dummyRoot.left = root.left;
                return dummyRoot.left;
            } else {
                TreeNode parent = root;
                TreeNode child = root.val > key ? root.left : root.right;
                while (child != null && child.val != key) {
                    parent = child;
                    child = child.val > key ? child.left : child.right;
                }
                if (child != null) {
                    boolean isLeftChild = parent.left == child;
                    if (child.left == null && child.right == null) {
                        if (isLeftChild)
                            parent.left = null;
                        else
                            parent.right = null;
                    } else if (child.left == null) {
                        if (isLeftChild)
                            parent.left = child.right;
                        else
                            parent.right = child.right;
                    } else if (child.right == null) {
                        if (isLeftChild)
                            parent.left = child.left;
                        else
                            parent.right = child.left;
                    } else {
                        TreeNode temp1 = child.right;
                        if (temp1.left == null) {
                            child.val = temp1.val;
                            child.right = temp1.right;
                        } else {
                            TreeNode temp2 = temp1.left;
                            while (temp2.left != null) {
                                temp2 = temp2.left;
                                temp1 = temp1.left;
                            }
                            child.val = temp2.val;
                            temp1.left = temp2.right;
                        }
                    }
                }
                return root;
            }
        }
    }
    
  • // OJ: https://leetcode.com/problems/delete-node-in-a-bst/
    // Time: O(H)
    // Space: O(H)
    class Solution {
    public:
        TreeNode* deleteNode(TreeNode* root, int key) {
            if (!root) 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) {
                auto p = root->left;
                while (p->right) p = p->right;
                root->val = p->val;
                root->left = deleteNode(root->left, root->val);
            } else if (root->right) {
                auto p = root->right;
                while (p->left) p = p->left;
                root->val = p->val;
                root->right = deleteNode(root->right, root->val);
            } else {
                delete root;
                root = NULL;
            }
            return root;
        }
    };
    
  • # 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
        """
    
        def delete(root, pre):
          if root.right:
            p = root.right
            while p.left:
              p = p.left
            p.left = root.left
          if root is pre.left:
            pre.left = root.right or root.left
          if root is pre.right:
            pre.right = root.right or root.left
          root.left = None
    
        if not root:
          return root
        pre = dummy = TreeNode(float("inf"))
        dummy.left = root
        p = dummy
        while p:
          if key > p.val:
            pre = p
            p = p.right
          elif key < p.val:
            pre = p
            p = p.left
          else:
            delete(p, pre)
            break
        return dummy.left
    
    

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